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    NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

    Edited By Ramraj Saini | Updated on Sep 19, 2023 09:27 AM IST

    NCERT Three Dimensional Geometry Class 12 Questions And Answers

    NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry - In the Class 12 Maths Chapter 11 NCERT solutions you will learn how to use vector algebra to study three dimensional geometry. Understanding vector algebra is a prerequisite for 3d geometry class 12. The practice of three dimensional geometry class 12 solutions will be helpful in solving three dimensional problems given in NCERT Books for Class 12.

    NCERT Class 12 Maths solutions chapter 11 makes the study simple and effective. It is also very helpful to solve problems asked in the CBSE Board as well as entrance exam related to chapter three dimensional geometry class 12. NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry will build your base for many other higher-level concepts like tensors and manifolds due to that this ch 11maths class 12 becomes very important. If you want to know more about 3d geometry class 12 then you can also check NCERT solutions for other classes.

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    NCERT Three Dimensional Geometry Class 12 Questions And Answers PDF Free Download

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    NCERT Class 12 Maths Chapter 11 Question Answer - Important Formulae

    >> Distance Formula:

    The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by:

    AB = √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²]
    The distance between a point A(x, y, z) and the origin O(0, 0, 0) is given by:

    OA = √(x² + y² + z²)

    >> Section Formula: The coordinates of the point R, which divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally or externally in the ratio m:n, are given by:

    Internal Division: (mx2 + nx1) / (m + n), (my2 + ny1) / (m + n), (mz2 + nz1) / (m + n)
    External Division: (mx2 - nx1) / (m - n), (my2 - ny1) / (m - n), (mz2 - nz1) / (m - n)

    >> Midpoint Formula: The coordinates of the mid-point of the line segment joining (x1, y1) and (x2, y2) are:

    [(x1 + x2) / 2, (y1 + y2) / 2]

    >> Coordinates of Centroid of a Triangle: Given vertices (x1, y1), (x2, y2), and (x3, y3) of a triangle, the coordinates of the centroid are:

    [(x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3]

    >> Incentre of a Triangle: Given vertices (x1, y1), (x2, y2), and (x3, y3) of a triangle, the coordinates of the incenter are:

    [(ax1 + bx2 + cx3) / (a + b + c), (ay1 + by2 + cy3) / (a + b + c)]

    >> Centroid of a Tetrahedron: Given vertices (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), and (x4, y4, z4) of a tetrahedron, the coordinates of the centroid are:

    [(x1 + x2 + x3 + x4) / 4, (y1 + y2 + y3 + y4) / 4, (z1 + z2 + z3 + z4) / 4]

    >> Direction Cosines of a Line: If a directed line OP makes angles α, β, and γ with the positive X-axis, Y-axis, and Z-axis, respectively, then the direction cosines l, m, and n are:

    l = cos α, m = cos β, n = cos γ Also, the sum of squares of direction cosines is always 1:

    l² + m² + n² = 1

    >> Direction Ratios of a Line: Direction ratios of a line are denoted as a, b, and c. They are proportional to the direction cosines:

    l/a = m/b = n/c

    >> Angle between Two Line Segments: If a1, b1, c1, and a2, b2, c2 are the direction ratios of two lines and θ is the acute angle between them, then:

    cos θ = |(a1a2 + b1b2 + c1c2) / (√(a1² + b1² + c1²) √(a2² + b2² + c2²))|

    >> Perpendicular and Parallel Lines: Two lines are perpendicular if: a1a2 + b1b2 + c1c2 = 0

    Two lines are parallel if: a1/a2 = b1/b2 = c1/c2

    >> Projection of a Line Segment on a Line: Given points P(x1, y1, z1) and Q(x2, y2, z2) and a line with direction cosines l, m, n, the projection of PQ on the line is:

    |l(x2 - x1) + m(y2 - y1) + n(z2 - z1)|

    >> Equation of a Plane: A plane in 3-D space can be represented in various forms:

    • General form: ax + by + cz + d = 0 (where a, b, c are not all zero)

    • Normal form: lx + my + nz = p

    • Plane through a point (x1, y1, z1): a(x - x1) + b(y - y1) + c(z - z1) = 0

    • Intercept form: (x/a) + (y/b) + (z/c) = 1

    • Vector form: (r- a).n = 0 or r.n = a.n

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    >> Planes Parallel to Axes: Planes parallel to the X-axis, Y-axis, and Z-axis are represented as:

    • Plane Parallel to X-axis: by + cz + d = 0

    • Plane Parallel to Y-axis: ax + cz + d = 0

    • Plane Parallel to Z-axis: ax + by + d = 0

    Free download NCERT Class 12 Maths Chapter 11 Question Answer for CBSE Exam.

    NCERT Three Dimensional Geometry Class 12 Questions And Answers (Intext Questions and Exercise)

    NCERT Class 12 Maths Chapter 11 Question Answer - Exercise: 11.1

    Question:1 If a line makes angles 90^{\circ}, 135^{\circ},45^{\circ} with the x, y and z-axes respectively, find its direction cosines.

    Answer:

    Let the direction cosines of the line be l,m, and n.

    So, we have

    l = \cos90^{\circ}=0

    m = \cos135^{\circ}=-\frac{1}{\sqrt2}

    n= \cos45^{\circ}=\frac{1}{\sqrt2}

    Therefore the direction cosines of the lines are 0,\ -\frac{1}{\sqrt2},and\ \ \frac{1}{\sqrt2} .

    Question:2 Find the direction cosines of a line which makes equal angles with the coordinate axes.

    Answer:

    If the line is making equal angle with the coordinate axes. Then,

    Let the common angle made is \alpha with each coordinate axes.

    Therefore, we can write;

    l = \cos \alpha,\ m= \cos \alpha,and\ n= \cos \alpha

    And as we know the relation; l^2+m^2+n^2 = 1

    \Rightarrow \cos^2 \alpha +\cos^2 \alpha+\cos^2 \alpha = 1

    \Rightarrow \cos^2 \alpha = \frac{1}{3}

    or \cos \alpha =\pm \frac{1}{\sqrt3}

    Thus the direction cosines of the line are \pm \frac{1}{\sqrt3},\ \pm \frac{1}{\sqrt3},and\ \pm \frac{1}{\sqrt3}

    Question:3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

    Answer:

    GIven a line has direction ratios of -18, 12, – 4 then its direction cosines are;

    Line having direction ratio -18 has direction cosine:

    \frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{-18}{22} = \frac{-9}{11}

    Line having direction ratio 12 has direction cosine:

    \frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{12}{22} =\frac{6}{11}

    Line having direction ratio -4 has direction cosine:

    \frac{12}{\sqrt{(-4)^2+(12)^2+(-4)^2}} = \frac{-4}{22} = \frac{-2}{11}

    Thus, the direction cosines are \frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11} .

    Question:4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.

    Answer:

    We have the points, A (2, 3, 4),B (– 1, – 2, 1),C (5, 8, 7);

    And as we can find the direction ratios of the line joining the points (x_{1},y_{1},z_{1}) \ and\ (x_{2},y_{2},z_{2}) is given by x_{2}-x_{1}, y_{2}-y_{1}, \ and\ z_{2}-z_{1}.

    The direction ratios of AB are (-1-2), (-2-3),\ and\ (1-4) i.e., -3,\ -5,\ and\ -3

    The direction ratios of BC are (5-(-1)), (8-(-2)),\ and\ (7-1) i.e., 6,\ 10,\ and\ 6 .

    We can see that the direction ratios of AB and BC are proportional to each other and is -2 times.

    \therefore AB is parallel to BC. and as point B is common to both AB and BC,

    Hence the points A, B and C are collinear.

    Question:5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

    Answer:

    Given vertices of the triangle \triangle ABC (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

    1633927872313

    Finding each side direction ratios;

    \Rightarrow Direction ratios of side AB are (-1-3), (1-5),\ and\ (2-(-4)) i.e.,

    -4,-4,\ and\ 6.

    Therefore its direction cosines values are;

    \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}

    SImilarly for side BC;

    \Rightarrow Direction ratios of side BC are (-5-(-1)), (-5-1),\ and\ (-2-2) i.e.,

    -4,-6,\ and\ -4.

    Therefore its direction cosines values are;

    \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}

    \Rightarrow Direction ratios of side CA are (-5-3), (-5-5),\ and\ (-2-(-4)) i.e.,

    -8,-10,\ and\ 2.

    Therefore its direction cosines values are;

    \frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}} or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}

    NCERT Class 12 Maths Chapter 11 Question Answer - Exercise: 11.2

    Question:1 Show that the three lines with direction cosines

    \frac{12}{13}, \frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13} are mutually perpendicular.

    Answer:

    GIven direction cosines of the three lines;

    L_{1}\ \left ( \frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \right ) L_{2}\ \left ( \frac{4}{13}, \frac{12}{13},\frac{3}{13} \right ) L_{3}\ \left ( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \right )

    And we know that two lines with direction cosines l_{1},m_{1},n_{1} and l_{2},m_{2},n_{2} are perpendicular to each other, if l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0

    Hence we will check each pair of lines:

    Lines L_{1}\ and\ L_{2} ;

    l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{12}{13}\times\frac{4}{13} \right ]+\left [ \frac{-3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times \frac{3}{13} \right ]

    = \left [ \frac{48}{169} \right ]-\left [ \frac{36}{169} \right ]-\left [ \frac{12}{169} \right ]= 0

    \therefore the lines L_{1}\ and\ L_{2} are perpendicular.

    Lines L_{2}\ and\ L_{3} ;

    l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{4}{13}\times\frac{3}{13} \right ]+\left [ \frac{12}{13}\times\frac{-4}{13} \right ]+\left [ \frac{3}{13}\times \frac{12}{13} \right ]

    = \left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]+\left [ \frac{36}{169} \right ]= 0

    \therefore the lines L_{2}\ and\ L_{3} are perpendicular.

    Lines L_{3}\ and\ L_{1} ;

    l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times\frac{-3}{13} \right ]+\left [ \frac{12}{13}\times \frac{-4}{13} \right ]

    = \left [ \frac{36}{169} \right ]+\left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]= 0

    \therefore the lines L_{3}\ and\ L_{1} are perpendicular.

    Thus, we have all lines are mutually perpendicular to each other.

    Question:2 Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

    Answer:

    We have given points where the line is passing through it;

    Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and line joining the points (0, 3, 2) and (3, 5, 6).is CD.

    So, we will find the direction ratios of the lines AB and CD;

    Direction ratios of AB are a_{1},b_{1}, c_{1}

    (3-1),\ (4-(-1)),\ and\ (-2-2) or 2,\ 5,\ and\ -4

    Direction ratios of CD are a_{2},b_{2}, c_{2}

    (3-0),\ (5-3)),\ and\ (6-2) or 3,\ 2,\ and\ 4 .

    Now, lines AB and CD will be perpendicular to each other if a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =0

    a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =\left ( 2\times3 \right ) +\left ( 5\times2 \right )+ \left ( -4\times 4 \right )

    = 6+10-16 = 0

    Therefore, AB and CD are perpendicular to each other.

    Question:3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

    Answer:

    We have given points where the line is passing through it;

    Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and line joining the points (– 1, – 2, 1) and (1, 2, 5)..is CD.

    So, we will find the direction ratios of the lines AB and CD;

    Direction ratios of AB are a_{1},b_{1}, c_{1}

    (2-4),\ (3-7),\ and\ (4-8) or -2,\ -4,\ and\ -4

    Direction ratios of CD are a_{2},b_{2}, c_{2}

    (1-(-1)),\ (2-(-2)),\ and\ (5-1) or 2,\ 4,\ and\ 4 .

    Now, lines AB and CD will be parallel to each other if \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

    Therefore we have now;

    \frac{a_{1}}{a_{2}} = \frac{-2}{2}=-1 \frac{b_{1}}{b_{2}} = \frac{-4}{4}=-1 \frac{c_{1}}{c_{2}} = \frac{-4}{4}=-1

    \therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

    Hence we can say that AB is parallel to CD.

    Question:4 Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3\widehat{i}+2\widehat{j}-2\widehat{k} .

    Answer:

    It is given that the line is passing through A (1, 2, 3) and is parallel to the vector \vec{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}

    We can easily find the equation of the line which passes through the point A and is parallel to the vector \vec{b} by the known relation;

    \vec{r} = \vec{a} +\lambda\vec{b} , where \lambda is a constant.

    So, we have now,

    \\\mathrm{\Rightarrow \vec{r} = \widehat{i}+2\widehat{j}+3\widehat{k} + \lambda(3\widehat{i}+2\widehat{j}-2\widehat{k})}

    Thus the required equation of the line.

    Answer:

    Given that the line is passing through the point with position vector 2\widehat{i}-\widehat{j}+4\widehat{k} and is in the direction of the line \widehat{i}+2\widehat{j}-\widehat{k} .

    And we know the equation of the line which passes through the point with the position vector \vec{a} and parallel to the vector \vec{b} is given by the equation,

    \vec{r} = \vec{a} +\lambda\vec{b}

    \Rightarrow \vec{r} =2\widehat{i}-\widehat{j}+4\widehat{k} + \lambda(\widehat{i}+2\widehat{j}-\widehat{k})

    So, this is the required equation of the line in the vector form.

    \vec{r} =x\widehat{i}+y\widehat{j}+z\widehat{k} = (\lambda+2)\widehat{i}+(2\lambda-1)\widehat{j}+(-\lambda+4)\widehat{k}

    Eliminating \lambda , from the above equation we obtain the equation in the Cartesian form :

    \frac{x-2}{1}= \frac{y+1}{2} =\frac{z-4}{-1}

    Hence this is the required equation of the line in Cartesian form.

    Question:6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} .

    Answer:

    Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} ;

    The direction ratios of the line, \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} are 3,5 and 6 .

    So, the required line is parallel to the above line.

    Therefore we can take direction ratios of the required line as 3k , 5k , and 6k , where k is a non-zero constant.

    And we know that the equation of line passing through the point (x_{1},y_{1},z_{1}) and with direction ratios a, b, c is written by: \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c} .

    Therefore we have the equation of the required line:

    \frac{x+2}{3k} = \frac{y-4}{5k} = \frac{z+5}{6k}

    or \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} = k

    The required line equation.

    Question:7 The cartesian equation of a line is \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7} . Write its vector form .

    Answer:

    Given the Cartesian equation of the line;

    \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}

    Here the given line is passing through the point (5,-4,6) .

    So, we can write the position vector of this point as;

    \vec{a} = 5\widehat{i}-4\widehat{j}+6\widehat{k}

    And the direction ratios of the line are 3 , 7 , and 2.

    This implies that the given line is in the direction of the vector, \vec{b} = 3\widehat{i}+7\widehat{j}+2\widehat{k} .

    Now, we can easily find the required equation of line:

    As we know that the line passing through the position vector \vec{a} and in the direction of the vector \vec{b} is given by the relation,

    \vec{r} = \vec{a} + \lambda \vec{b},\ \lambda \epsilon R

    So, we get the equation.

    \vec{r} = 5\widehat{i}-4\widehat{j}+6\widehat{k} + \lambda(3\widehat{i}+7\widehat{j}+2\widehat{k}),\ \lambda \epsilon R

    This is the required equation of the line in the vector form.

    Question:8 Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).

    Answer:

    GIven that the line is passing through the (0,0,0) and (5,-2,3)

    Thus the required line passes through the origin.

    \therefore its position vector is given by,

    \vec{a} = \vec{0}

    So, the direction ratios of the line through (0,0,0) and (5,-2,3) are,

    (5-0) = 5, (-2-0) = -2, (3-0) = 3

    The line is parallel to the vector given by the equation, \vec{b} = 5\widehat{i}-2\widehat{j}+3\widehat{k}

    Therefore the equation of the line passing through the point with position vector \vec{a} and parallel to \vec{b} is given by;

    \vec{r} = \vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R

    \Rightarrow\vec{r} = 0+\lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})

    \Rightarrow\vec{r} = \lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})

    Now, the equation of the line through the point (x_{1},y_{1},z_{1}) and the direction ratios a, b, c is given by;

    \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}

    Therefore the equation of the required line in the Cartesian form will be;

    \frac{x-0}{5} = \frac{y-0}{-2} =\frac{z-0}{3}

    OR \frac{x}{5} = \frac{y}{-2} =\frac{z}{3}

    Question:9 Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

    Answer:

    Let the line passing through the points A(3,-2,-5) and B(3,-2,6) is AB;

    Then as AB passes through through A so, we can write its position vector as;

    \vec{a} =3\widehat{i}-2\widehat{j}-5\widehat{k}

    Then direction ratios of PQ are given by,

    (3-3)= 0,\ (-2+2) = 0,\ (6+5)=11

    Therefore the equation of the vector in the direction of AB is given by,

    \vec{b} =0\widehat{i}-0\widehat{j}+11\widehat{k} = 11\widehat{k}

    We have then the equation of line AB in vector form is given by,

    \vec{r} =\vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R

    \Rightarrow \vec{r} = (3\widehat{i}-2\widehat{j}-5\widehat{k}) + 11\lambda\widehat{k}

    So, the equation of AB in Cartesian form is;

    \frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}

    or \frac{x-3}{0} =\frac{y+2}{0} =\frac{z+5}{11}

    Question:10 Find the angle between the following pairs of lines:

    (i) \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and \overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

    Answer:

    To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

    \cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

    We have two lines :

    \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and

    \overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

    The given lines are parallel to the vectors \vec{b_{1}}\ and\ \vec{b_{2}} ;

    where \vec{b_{1}}= 3\widehat{i}+2\widehat{j}+6\widehat{k} and \vec{b_{2}}= \widehat{i}+2\widehat{j}+2\widehat{k} respectively,

    Then we have

    \vec{b_{1}}.\vec{b_{2}} =(3\widehat{i}+2\widehat{j}+6\widehat{k}).(\widehat{i}+2\widehat{j}+2\widehat{k})

    =3+4+12 = 19

    and |\vec{b_{1}}| = \sqrt{3^2+2^2+6^2} = 7

    |\vec{b_{2}}| = \sqrt{1^2+2^2+2^2} = 3

    Therefore we have;

    \cos A = \left | \frac{19}{7\times3} \right | = \frac{19}{21}

    or A = \cos^{-1} \left ( \frac{19}{21} \right )

    Question:10 Find the angle between the following pairs of lines:

    (ii) \overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and \overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

    Answer:

    To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

    \cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

    We have two lines :

    \overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and

    \overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

    The given lines are parallel to the vectors \vec{b_{1}}\ and\ \vec{b_{2}} ;

    where \vec{b_{1}}= \widehat{i}-\widehat{j}-2\widehat{k} and \vec{b_{2}}= 3\widehat{i}-5\widehat{j}-4\widehat{k} respectively,

    Then we have

    \vec{b_{1}}.\vec{b_{2}} =(\widehat{i}-\widehat{j}-2\widehat{k}).(3\widehat{i}-5\widehat{j}-4\widehat{k})

    =3+5+8 = 16

    and |\vec{b_{1}}| = \sqrt{1^2+(-1)^2+(-2)^2} = \sqrt{6}

    |\vec{b_{2}}| = \sqrt{3^2+(-5)^2+(-4)^2} = \sqrt{50} = 5\sqrt2

    Therefore we have;

    \cos A = \left | \frac{16}{\sqrt6 \times5\sqrt2} \right | = \frac{16}{10\sqrt3}

    or A = \cos^{-1} \left ( \frac{8}{5\sqrt3} \right )

    Question:11 Find the angle between the following pair of lines:

    (i) \frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3} and \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

    Answer:

    Given lines are;

    \frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3} and \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

    So, we two vectors \vec{b_{1}}\ and\ \vec{b_{2}} which are parallel to the pair of above lines respectively.

    \vec{b_{1}}\ =2\widehat{i}+5\widehat{j}-3\widehat{k} and \vec{b_{2}}\ =-\widehat{i}+8\widehat{j}+4\widehat{k}

    To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

    \cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

    Then we have

    \vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+5\widehat{j}-3\widehat{k}).(-\widehat{i}+8\widehat{j}+4\widehat{k})

    =-2+40-12 = 26

    and |\vec{b_{1}}| = \sqrt{2^2+5^2+(-3)^2} = \sqrt{38}

    |\vec{b_{2}}| = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9

    Therefore we have;

    \cos A = \left | \frac{26}{\sqrt{38} \times9} \right | = \frac{26}{9\sqrt{38}}

    or A = \cos^{-1} \left ( \frac{26}{9\sqrt{38}} \right )

    Question:11 Find the angle between the following pair of lines:

    (ii) \frac{x}{2}= \frac{y}{2}=\frac{z}{1} and \frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}

    Answer:

    Given lines are;

    \frac{x}{2}= \frac{y}{2}=\frac{z}{1} and \frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}

    So, we two vectors \vec{b_{1}}\ and\ \vec{b_{2}} which are parallel to the pair of above lines respectively.

    \vec{b_{1}}\ =2\widehat{i}+2\widehat{j}+\widehat{k} and \vec{b_{2}}\ =4\widehat{i}+\widehat{j}+8\widehat{k}

    To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

    \cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

    Then we have

    \vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+2\widehat{j}+\widehat{k}).(4\widehat{i}+\widehat{j}+8\widehat{k})

    =8+2+8 = 18

    and |\vec{b_{1}}| = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3

    |\vec{b_{2}}| = \sqrt{(4)^2+(1)^2+(8)^2} = \sqrt{81} = 9

    Therefore we have;

    \cos A = \left | \frac{18}{ 3\times9} \right | = \frac{2}{3}

    or A = \cos^{-1} \left ( \frac{2}{3} \right )

    Question:12 Find the values of p so that the lines \frac{1-x}{3}=\frac{7y-14}{2p}= \frac{z-3}{2} and \frac{7-7x}{3p}=\frac{y-5}{1}= \frac{6-z}{5} are at right angles.

    Answer:

    First we have to write the given equation of lines in the standard form;

    \frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}= \frac{z-3}{2} and \frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}= \frac{z-6}{-5}

    Then we have the direction ratios of the above lines as;

    -3,\ \frac{2p}{7},\ 2 and \frac{-3p}{7},\ 1,\ -5 respectively..

    Two lines with direction ratios a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2} are perpendicular to each other if, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0

    \therefore (-3).\left ( \frac{-3p}{7} \right )+(\frac{2p}{7}).(1) + 2.(-5) = 0

    \Rightarrow \frac{9p}{7}+ \frac{2p}{7} =10

    \Rightarrow 11p =70

    \Rightarrow p =\frac{70}{11}

    Thus, the value of p is \frac{70}{11} .

    Question:13 Show that the lines \frac{x-5}{7}=\frac{y+2}{-3}=\frac{z}{1} and \frac{x}{1}=\frac{y}{2}=\frac{z}{3} are perpendicular to each other.

    Answer:

    First, we have to write the given equation of lines in the standard form;

    \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} and \frac{x}{1}=\frac{y}{2}=\frac{z}{3}

    Then we have the direction ratios of the above lines as;

    7,\ -5,\ 1 and 1,\ 2,\ 3 respectively..

    Two lines with direction ratios a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2} are perpendicular to each other if, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0

    \therefore 7(1) + (-5)(2)+1(3) = 7-10+3 = 0

    Therefore the two lines are perpendicular to each other.

    Question:14 Find the shortest distance between the lines

    \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k}) and \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})

    Answer:

    So given equation of lines;

    \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k}) and \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k}) in the vector form.

    Now, we can find the shortest distance between the lines \vec{r} = \vec{a_{1}}+\lambda\vec{b_{1}} and \vec{r} = \vec{a_{2}}+\mu \vec{b_{2}} , is given by the formula,

    d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

    Now comparing the values from the equation, we obtain

    \vec{a_{1}} = \widehat{i}+2\widehat{j}+\widehat{k} \vec{b_{1}} = \widehat{i}-\widehat{j}+\widehat{k}

    \vec{a_{2}} = 2\widehat{i}-\widehat{j}-\widehat{k} \vec{b_{2}} = 2\widehat{i}+\widehat{j}+2\widehat{k}

    \vec{a_{2}} -\vec{a_{1}} =\left ( 2\widehat{i}-\widehat{j}-\widehat{k} \right ) - \left ( \widehat{i}+2\widehat{j}+\widehat{k} \right ) = \widehat{i}-3\widehat{j}-2\widehat{k}

    Then calculating

    \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 & -1 &1 \\ 2& 1 &2 \end{vmatrix}

    \vec{b_{1}}\times \vec{b_{2}} = (-2-1)\widehat{i} - (2-2) \widehat{j} +(1+2) \widehat{k} = -3\widehat{i}+3\widehat{k}

    \Rightarrow \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} =\sqrt{18} =3\sqrt2

    So, substituting the values now in the formula above we get;

    d =\left | \frac{\left ( -3\widehat{i}+3\widehat{k} \right ).(\widehat{i}-3\widehat{j}-2\widehat{k})}{3\sqrt2} \right |

    \Rightarrow d = \left | \frac{-3.1+3(-2)}{3\sqrt2} \right |

    d = \left | \frac{-9}{3\sqrt2} \right | = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}

    Therefore, the shortest distance between the two lines is \frac{3\sqrt2}{2} units.

    Question:15 Find the shortest distance between the lines

    \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

    Answer:

    We have given two lines:

    \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

    Calculating the shortest distance between the two lines,

    \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} and \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}

    by the formula

    d = \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2}}

    Now, comparing the given equations, we obtain

    x_{1} = -1,\ y_{1} =-1,\ z_{1} =-1

    a_{1} = 7,\ b_{1} =-6,\ c_{1} =1

    x_{2} = 3,\ y_{2} =5,\ z_{2} =7

    a_{2} = 1,\ b_{2} =-2,\ c_{2} =1

    Then calculating determinant

    \begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix} = \begin{vmatrix} 4 &6 &8 \\ 7& -6& 1\\ 1& -2& 1 \end{vmatrix}

    = 4(-6+2)-6(7-1)+8(-14+6)

    = -16-36-64

    =-116

    Now calculating the denominator,

    \sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2} = \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2} = \sqrt{16+36+64}

    = \sqrt{116} = 2\sqrt{29}

    So, we will substitute all the values in the formula above to obtain,

    d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = \frac{-2\times29}{\sqrt{29}} = -2\sqrt{29}

    Since distance is always non-negative, the distance between the given lines is

    2\sqrt{29} units.

    Question:16 Find the shortest distance between the lines whose vector equations are \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) and

    \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})

    Answer:

    Given two equations of line

    \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k}) in the vector form.

    So, we will apply the distance formula for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}} and \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

    d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

    After comparing the given equations, we obtain

    \vec{a_{1}} = \widehat{i}+2\widehat{j}+3\widehat{k} \vec{b_{1}} = \widehat{i}-3\widehat{j}+2\widehat{k}

    \vec{a_{2}} = 4\widehat{i}+5\widehat{j}+6\widehat{k} \vec{b_{2}} = 2\widehat{i}+3\widehat{j}+\widehat{k}

    \vec{a_{2}}-\vec{a_{1}} = (4\widehat{i}+5\widehat{j}+6\widehat{k}) - (\widehat{i}+2\widehat{j}+3\widehat{k})

    = 3\widehat{i}+3\widehat{j}+3\widehat{k}

    Then calculating the determinant value numerator.

    \vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1& -3 &2 \\ 2& 3& 1 \end{vmatrix}

    = (-3-6)\widehat{i}-(1-4)\widehat{j}+(3+6)\widehat{k} = -9\widehat{i}+3\widehat{j}+9\widehat{k}

    That implies, \left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(-9)^2+(3)^2+(9)^2}

    = \sqrt{81+9+81} = \sqrt{171} =3\sqrt{19}

    \left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(-9\widehat{i}+3\widehat{j}+9\widehat{k})(3\widehat{i}+3\widehat{j}+3\widehat{k})

    = (-9\times3)+(3\times3)+(9\times3) = 9

    Now, after substituting the value in the above formula we get,

    d= \left | \frac{9}{3\sqrt{19}} \right | = \frac{3}{\sqrt{19}}

    Therefore, \frac{3}{\sqrt{19}} is the shortest distance between the two given lines.

    Question:17 Find the shortest distance between the lines whose vector equations are

    \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k} and \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}

    Answer:

    Given two equations of the line

    \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k} \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k} in the vector form.

    So, we will apply the distance formula for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}} and \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

    d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

    After comparing the given equations, we obtain

    \vec{a_{1}} = \widehat{i}-2\widehat{j}+3\widehat{k} \vec{b_{1}} = -\widehat{i}+\widehat{j}-2\widehat{k}

    \vec{a_{2}} = \widehat{i}-\widehat{j}-\widehat{k} \vec{b_{2}} = \widehat{i}+2\widehat{j}-2\widehat{k}

    \vec{a_{2}}-\vec{a_{1}} = (\widehat{i}-\widehat{j}-\widehat{k}) - (\widehat{i}-2\widehat{j}+3\widehat{k}) = \widehat{j}-4\widehat{k}

    Then calculating the determinant value numerator.

    \vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ -1& 1 &-2 \\ 1& 2& -2 \end{vmatrix}

    = (-2+4)\widehat{i}-(2+2)\widehat{j}+(-2-1)\widehat{k} = 2\widehat{i}-4\widehat{j}-3\widehat{k}

    That implies,

    \left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(2)^2+(-4)^2+(-3)^2}

    = \sqrt{4+16+9} = \sqrt{29}

    \left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(2\widehat{i}-4\widehat{j}-3\widehat{k})(\widehat{j}-4\widehat{k}) = -4+12 = 8

    Now, after substituting the value in the above formula we get,

    d= \left | \frac{8}{\sqrt{29}} \right | = \frac{8}{\sqrt{29}}

    Therefore, \frac{8}{\sqrt{29}} units are the shortest distance between the two given lines.


    NCERT class 12 three dimensional geometry ncert solutions - Exercise: 11.3

    Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    z = 2

    Answer:

    Equation of plane Z=2, i.e. 0x+0y+z=2

    The direction ratio of normal is 0,0,1

    \therefore \, \, \, \sqrt{0^2+0^2+1^2}=1

    Divide equation 0x+0y+z=2 by 1 from both side

    We get, 0x+0y+z=2

    Hence, direction cosins are 0,0,1.

    The distance of the plane from the origin is 2.

    Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    x + y + z = 1

    Answer:

    Given the equation of the plane is x+y+z=1 or we can write 1x+1y+1z=1

    So, the direction ratios of normal from the above equation are, 1,\1,\ and\ 1 .

    Therefore \sqrt{1^2+1^2+1^2} =\sqrt{3}

    Then dividing both sides of the plane equation by \sqrt{3} , we get

    \frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3}=\frac{1}{\sqrt3}

    So, this is the form of lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

    \therefore The direction cosines of the given line are \frac{1}{\sqrt3},\ \frac{1}{\sqrt3},\ \frac{1}{\sqrt3} and the distance of the plane from the origin is \frac{1}{\sqrt3} units.

    Question:1(c) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    2x + 3y - z = 5

    Answer:

    Given the equation of plane is 2x+3y-z=5

    So, the direction ratios of normal from the above equation are, 2,\3,\ and\ -1 .

    Therefore \sqrt{2^2+3^2+(-1)^2} =\sqrt{14}

    Then dividing both sides of the plane equation by \sqrt{14} , we get

    \frac{2x}{\sqrt{14}}+\frac{3y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}

    So, this is the form of lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

    \therefore The direction cosines of the given line are \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-1}{\sqrt{14}} and the distance of the plane from the origin is \frac{5}{\sqrt{14}} units.

    Question:1(d) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    5y + 8 = 0

    Answer:

    Given the equation of plane is 5y+8=0 or we can write 0x-5y+0z=8

    So, the direction ratios of normal from the above equation are, 0,\ -5,\ and\ 0 .

    Therefore \sqrt{0^2+(-5)^2+0^2} =5

    Then dividing both sides of the plane equation by 5 , we get

    -y = \frac{8}{5}

    So, this is the form of lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

    \therefore The direction cosines of the given line are 0,\ -1,\ and\ 0 and the distance of the plane from the origin is \frac{8}{5} units.

    Question:2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3\widehat{i}+5\widehat{j}-6\widehat{k} .

    Answer:

    We have given the distance between the plane and origin equal to 7 units and normal to the vector 3\widehat{i}+5\widehat{j}-6\widehat{k} .

    So, it is known that the equation of the plane with position vector \vec{r} is given by, the relation,

    \vec{r}.\widehat{n} =d , where d is the distance of the plane from the origin.

    Calculating \widehat{n} ;

    \widehat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{(3)^2+(5)^2+(6)^2}} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}}

    \vec{r}.\left ( \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}} \right ) = 7 is the vector equation of the required plane.

    Question:3(a) Find the Cartesian equation of the following planes:

    \overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2

    Answer:

    Given the equation of the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2

    So we have to find the Cartesian equation,

    Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,

    \vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

    Hence we have,

    (x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}-\widehat{k}) =2

    Or, x+y-z=2

    Therefore this is the required Cartesian equation of the plane.

    Question:3(b) Find the Cartesian equation of the following planes:

    \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

    Answer:

    Given the equation of plane \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

    So we have to find the Cartesian equation,

    Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,

    \vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

    Hence we have,

    (x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}+3\widehat{j}-4\widehat{k}) =1

    Or, 2x+3y-4z=1

    Therefore this is the required Cartesian equation of the plane.

    Question:3(c) Find the Cartesian equation of the following planes:

    \overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15

    Answer:

    Given the equation of plane \overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15

    So we have to find the Cartesian equation,

    Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by, \vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

    Hence we have,

    (x\widehat{i}+y\widehat{j}+z\widehat{k}).\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right] =15

    Or, (s-2t)x+(3-t)y+(2s+t)z=15

    Therefore this is the required Cartesian equation of the plane.

    Question:4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

    2 x + 3y + 4 z - 12 = 0

    Answer:

    Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

    Given a plane equation 2x+3y+4z-12=0 ,

    Or, 2x+3y+4z=12

    The direction ratios of the normal of the plane are 2, 3 and 4 .

    Therefore \sqrt{(2)^2+(3)^2+(4)^2} = \sqrt{29}

    So, now dividing both sides of the equation by \sqrt{29} we will obtain,

    \frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}

    This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

    Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd) .

    \therefore The coordinates of the foot of the perpendicular are;

    \left [ \frac{2}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{3}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{4}{\sqrt{29}}.\frac{12}{\sqrt{29}} \right ] or \left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]

    Question:4(b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

    3y + 4z - 6 = 0

    Answer:

    Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

    Given a plane equation 3y+4z-6=0 ,

    Or, 0x+3y+4z=6

    The direction ratios of the normal of the plane are 0,3 and 4 .

    Therefore \sqrt{(0)^2+(3)^2+(4)^2} = 5

    So, now dividing both sides of the equation by 5 we will obtain,

    0x+\frac{3}{5}y+\frac{4}{5}z = \frac{6}{5}

    This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

    Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd) .

    \therefore The coordinates of the foot of the perpendicular are;

    \left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right ) or \left ( 0, \frac{18}{25}, \frac{24}{25} \right )

    Question:4(c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

    x + y + z = 1

    Answer:

    Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

    Given plane equation x+y+z=1 .

    The direction ratios of the normal of the plane are 1,1 and 1 .

    Therefore \sqrt{(1)^2+(1)^2+(1)^2} = \sqrt3

    So, now dividing both sides of the equation by \sqrt3 we will obtain,

    \frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{1}{\sqrt3}

    This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

    Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd) .

    \therefore The coordinates of the foot of the perpendicular are;

    \left ( \frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3} \right ) or \left ( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right ) ..

    Question: 4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

    5y + 8 = 0

    Answer:

    Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

    Given plane equation 5y+8=0 .

    or written as 0x-5y+0z=8

    The direction ratios of the normal of the plane are 0, -5 and 0 .

    Therefore \sqrt{(0)^2+(-5)^2+(0)^2} = 5

    So, now dividing both sides of the equation by 5 we will obtain,

    -y=\frac{8}{5}

    This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

    Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd) .

    \therefore The coordinates of the foot of the perpendicular are;

    \left ( 0,-1(\frac{8}{5}),0 \right ) or \left ( 0,\frac{-8}{5},0 \right ) .

    Question:5(a) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is \widehat{i}+\widehat{j}-\widehat{k}.

    Answer:

    Given the point A (1,0,-2) and the normal vector \widehat{n} which is perpendicular to the plane is \widehat{n} = \widehat{i}+\widehat{j}-\widehat{k}

    The position vector of point A is \vec {a} = \widehat{i}-2\widehat{k}

    So, the vector equation of the plane would be given by,

    (\vec{r}-\vec{a}).\widehat{n} = 0

    Or \left [ \vec{r}-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

    where \vec{r} is the position vector of any arbitrary point A(x,y,z) in the plane.

    \therefore \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}

    Therefore, the equation we get,

    \left [(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

    \Rightarrow \left [(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

    \Rightarrow(x-1)+y-(z+2) = 0

    \Rightarrow x+y-z-3=0 or x+y-z=3

    So, this is the required Cartesian equation of the plane.

    Question:5(b) Find the vector and cartesian equations of the planes

    that passes through the point (1,4, 6) and the normal vector to the plane is \widehat{i}-2\widehat{j}+\widehat{k} .

    Answer:

    Given the point A (1,4,6) and the normal vector \widehat{n} which is perpendicular to the plane is \widehat{n} = \widehat{i}-2\widehat{j}+\widehat{k}

    The position vector of point A is \vec {a} = \widehat{i}+4\widehat{j}+6\widehat{k}

    So, the vector equation of the plane would be given by,

    (\vec{r}-\vec{a}).\widehat{n} = 0

    Or \left [ \vec{r}-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

    where \vec{r} is the position vector of any arbitrary point A(x,y,z) in the plane.

    \therefore \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}

    Therefore, the equation we get,

    \left [ (x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

    \Rightarrow \left [(x-1)\widehat{i}+(y-4)\widehat{j}+(z-6)\widehat{k}\right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

    (x-1)-2(y-4)+(z-6)=0

    \Rightarrow x-2y+z+1=0

    So, this is the required Cartesian equation of the plane.

    Question:6(a) Find the equations of the planes that passes through three points.

    (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

    Answer:

    The equation of the plane which passes through the three points A(1,1,-1),\ B(6,4,-5),\ and\ C(-4,-2,3) is given by;

    Determinant method,

    \begin{vmatrix} 1 &1 &-1 \\ 6& 4 & -5\\ -4& -2 &3 \end{vmatrix} = (12-10)-(18-20)-(-12+16)

    Or, = 2+2-4 = 0

    Here, these three points A, B, C are collinear points.

    Hence there will be an infinite number of planes possible which passing through the given points.

    Question:6(b) Find the equations of the planes that passes through three points.

    (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

    Answer:

    The equation of the plane which passes through the three points A(1,1,0),\ B(1,2,1),\ and\ C(-2,2,-1) is given by;

    Determinant method,

    \begin{vmatrix} 1 &1 &0 \\ 1& 2 & 1\\ -2& 2 &-1 \end{vmatrix} = (-2-2)-(2+2)= -8 \neq 0

    As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

    Finding the equation of the plane through the points, (x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})\ and\ (x_{3},y_{3},z_{3})

    \begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix} = 0

    After substituting the values in the determinant we get,

    \begin{vmatrix} x-1 &y-1 &z \\ 0& 1 &1 \\ -3& 1&-1 \end{vmatrix} = 0

    \Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3) = 0

    \Rightarrow-2x+2-3y+3+3z = 0

    2x+3y-3z = 5

    So, this is the required Cartesian equation of the plane.

    Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.

    Answer:

    Given plane 2x + y-z = 5

    We have to find the intercepts that this plane would make so,

    Making it look like intercept form first:

    By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

    \frac{2}{5}x+\frac{y}{5}-\frac{z}{5} =1

    \Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5} =1

    So, as we know that from the equation of a plane in intercept form, \frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1 where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

    Therefore after comparison, we get the values of a,b, and c.

    a = \frac{5}{2},\ b=5,\ and\ c=-5 .

    Hence the intercepts are \frac{5}{2},\ 5,\ and\ -5 .

    Question:8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

    Answer:

    Given that the plane is parallel to the ZOX plane.

    So, we have the equation of plane ZOX as y = 0 .

    And an intercept of 3 on the y-axis \Rightarrow b =3

    Intercept form of a plane given by;

    \frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1

    So, here the plane would be parallel to the x and z-axes both.

    we have any plane parallel to it is of the form, y=a .

    Equation of the plane required is y=3 .

    Question:9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

    Answer:

    The equation of any plane through the intersection of the planes,

    3x-y+2z-4=0\ and\ x+y+z-2=0

    Can be written in the form of; (3x-y+2z-4)\ +\alpha( x+y+z-2)= 0 , where \alpha \epsilon R

    So, the plane passes through the point (2,2,1) , will satisfy the above equation.

    (3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0

    That implies 2+3\alpha= 0

    \alpha = \frac{-2}{3}

    Now, substituting the value of \alpha in the equation above we get the final equation of the plane;

    (3x-y+2z-4)\ +\alpha( x+y+z-2)= 0

    (3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0

    \Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0

    \Rightarrow 7x-5y+4z-8= 0 is the required equation of the plane.

    Question:10 Find the vector equation of the plane passing through the intersection of the planes \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})=7 , \overrightarrow{r}(2\widehat{i}+5\widehat{j}+3\widehat{k})=9 and through the point (2, 1, 3).

    Answer:

    Here \vec{n_{1}} =2 \widehat{i}+2\widehat{j}-3\widehat{k} and \vec{n_{2}} = 2\widehat{i}+5\widehat{j}+3\widehat{k}

    and d_{1} = 7 and d_{2} = 9

    Hence, using the relation \vec{r}.(\vec{n_{1}}+\lambda\vec{n_{2}}) = d_{1}+\lambda d_{2} , we get

    \vec{r}.[2\widehat{i}+2\widehat{j}-3\widehat{k}+\lambda(2\widehat{i}+5\widehat{j}+3\widehat{k})] = 7+9\lambda

    or \vec{r}.[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda ..............(1)

    where, \lambda is some real number.

    Taking \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k} , we get

    (\vec{x\widehat{i}+y\widehat{j}+z\widehat{k}}).[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda

    or x(2+2\lambda) + y(2+5\lambda) +z(3\lambda-3) = 7+9\lambda

    or 2x+2y-3z-7 + \lambda(2x+5y+3z-9) = 0 .............(2)

    Given that the plane passes through the point (2,1,3) , it must satisfy (2), i.e.,

    (4+2-9-7) + \lambda(4+5+9-9) = 0

    or \lambda = \frac{10}{9}

    Putting the values of \lambda in (1), we get

    \vec{r}\left [\left ( 2+\frac{20}{9} \right )\widehat{i}+\left ( 2+\frac{50}{9} \right )\widehat{j}+\left ( \frac{10}{3}-3 \right )\widehat{k} \right ] = 7+10

    or \vec{r}\left ( \frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{1}{3}\widehat{k} \right ) = 17

    or \vec{r}.\left ( 38\widehat{i}+68\widehat{j}+3\widehat{k} \right ) = 153

    which is the required vector equation of the plane.

    Question:11 Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

    Answer:

    The equation of the plane through the intersection of the given two planes, x+y+z =1 and 2x+3y+4z =5 is given in Cartesian form as;

    (x+y+z-1) +\lambda(2x+3y+4z -5) = 0

    or (1+2\lambda)x(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) = 0 ..................(1)

    So, the direction ratios of (1) plane are a_{1},b_{1},c_{1} which are (1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda) .

    Then, the plane in equation (1) is perpendicular to x-y+z= 0 whose direction ratios a_{2},b_{2},c_{2} are 1,-1,\ and\ 1 .

    As planes are perpendicular then,

    a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    we get,

    (1+2\lambda) -(1+3\lambda)+(1+4\lambda) = 0

    or 1+3\lambda = 0

    or \lambda = -\frac{1}{3}

    Then we will substitute the values of \lambda in the equation (1), we get

    \frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0

    or x-z+2=0

    This is the required equation of the plane.

    Question:12 Find the angle between the planes whose vector equations are \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3 .

    Answer:

    Given two vector equations of plane

    \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3 .

    Here, \vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k} and \vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}

    The formula for finding the angle between two planes,

    \cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right | .............................(1)

    \vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15

    |\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}

    and |\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}

    Now, we can substitute the values in the angle formula (1) to get,

    \cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |

    or \cos A =\frac{15}{\sqrt{731}}

    or A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )

    Question:13(a) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

    7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

    Answer:

    Two planes

    L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2} ,

    are said to Parallel:

    If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    and Perpendicular:

    If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    And the angle between L_{1}\ and\ L_{2} is given by the relation,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
    So, given two planes 7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0

    Here,

    a_{1} = 7,b_{1} = 5, c_{1} = 6 and a_{2} = 3,b_{2} = -1, c_{2} = -10

    So, applying each condition to check:

    Parallel check: \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    \Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}

    Clearly, the given planes are NOT parallel. \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

    Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    \Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0 .

    Clearly, the given planes are NOT perpendicular.

    Then find the angle between them,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |

    = \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |

    = \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |

    = \cos^{-1}\left ( \frac{44}{110} \right )

    = \cos^{-1}\left ( \frac{2}{5} \right )

    Question:13(b) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

    2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

    Answer:

    Two planes

    L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2} ,

    are said to Parallel:

    If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    and Perpendicular:

    If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    And the angle between L_{1}\ and\ L_{2} is given by the relation,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
    So, given two planes 2x + y + 3z -2 = 0\ and\ x -2y + 5 = 0

    Here,

    a_{1} = 2,b_{1} = 1, c_{1} = 3 and a_{2} = 1,b_{2} = -2, c_{2} = 0

    So, applying each condition to check:

    Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    \Rightarrow 2(1)+1(-2)+3(0) = 2-2+0 = 0 .

    Thus, the given planes are perpendicular to each other.

    Question:13(c) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

    2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

    Answer:

    Two planes

    L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2} ,

    are said to Parallel:

    If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    and Perpendicular:

    If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    And the angle between L_{1}\ and\ L_{2} is given by the relation,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
    So, given two planes 2x - 2y + 4z + 5 = 0\ and\ 3x -3y +6z -1 = 0

    Here,

    a_{1} = 2,b_{1} = -2, c_{1} = 4 and a_{2} = 3,b_{2} = -3, c_{2} = 6

    So, applying each condition to check:

    Parallel check: \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    \Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3},\ and\ \frac{c_{1}}{c_{2}} = \frac{4}{6}=\frac{2}{3}

    Thus, the given planes are parallel as \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

    Question:13(d) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

    2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

    Answer:

    Two planes

    L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2} ,

    are said to Parallel:

    If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    and Perpendicular:

    If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    And the angle between L_{1}\ and\ L_{2} is given by the relation,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
    So, given two planes 2x - y + 3z -1 = 0\ and\ 2x -y +3z + 3 = 0

    Here,

    a_{1} = 2,b_{1} = -1, c_{1} = 3 and a_{2} = 2,b_{2} = -1, c_{2} = 3

    So, applying each condition to check:

    Parallel check: \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    \Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1} =1,\frac{c_{1}}{c_{2}} = \frac{3}{3} = 1

    Therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

    Thus, the given planes are parallel to each other.

    Question:13(e) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

    4x + 8y + z – 8 = 0 and y + z – 4 = 0

    Answer:

    Two planes

    L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2} ,

    are said to Parallel:

    If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    and Perpendicular:

    If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    And the angle between L_{1}\ and\ L_{2} is given by the relation,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
    So, given two planes 4x + 8y + z -8 = 0\ and\ y + z - 4 = 0

    Here,

    a_{1} = 4,b_{1} = 8, c_{1} = 1 and a_{2} = 0,b_{2} = 1, c_{2} = 1

    So, applying each condition to check:

    Parallel check: \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

    \Rightarrow \frac{a_{1}}{a_{2}} =\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1},\frac{c_{1}}{c_{2}} = \frac{1}{1}

    Clearly, the given planes are NOT parallel as \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} .

    Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    \Rightarrow 4(0)+8(1)+1(1) =0+8+1 = 9 \neq 0 .

    Clearly, the given planes are NOT perpendicular.

    Then finding the angle between them,

    A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |

    = \cos^{-1}\left | \frac{9}{\sqrt{4^2+8^2+1^2}.\sqrt{0^2+1^2+1^2}} \right |

    = \cos^{-1}\left | \frac{9}{\sqrt{81}.\sqrt{2}} \right |

    = \cos^{-1}\left ( \frac{9}{9\sqrt{2}} \right ) = \cos^{-1}\left ( \frac{1}{\sqrt{2}} \right )

    = 45^{\circ}

    Question:14 In the following cases, find the distance of each of the given points from the corresponding given plane


    Answer:

    We know that the distance between a point P (x_{1},y_{1},z_{1}) and a plane Ax+By+Cz =D is given by,

    d =\left | \frac{Ax_{1}+By_{1}+Cz_{1}-D}{\sqrt{A^2+B^2+C^2}} \right | .......................(1)

    So, calculating for each case;

    (a) Point (0,0,0) and Plane 3x-4y+12z = 3

    Therefore, d =\left | \frac{3(0)-4(0)+12(0)-3}{\sqrt{3^2+(-4)^2+12^2}} \right | = \frac{3}{\sqrt{169}} = \frac{3}{13}

    (b) Point (3,-2,1) and Plane 2x-y+2z +3= 0

    Therefore, d =\left | \frac{2(3)-(-2)+2(1)+3}{\sqrt{2^2+(-1)^2+2^2}} \right | = \frac{13}{3}

    (c) Point (2,3,-5) and Plane x+2y-2z =9

    Therefore, d =\left | \frac{2+2(3)-2(-5)-9}{\sqrt{1^2+2^2+(-2)^2}} \right | = \frac{9}{3} = 3

    (d) Point (-6,0,0) and Plane 2x-3y+6z -2= 0

    Therefore, d =\left | \frac{2(-6)-3(0)+6(0)-2}{\sqrt{2^2+(-3)^2+6^2}} \right | = \frac{-14}{\sqrt{49}} = \frac{14}{7} =2


    NCERT class 12 three dimensional geometry ncert solutions - Miscellaneous Exercise

    Question:1 Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

    Answer:

    We can assume the line joining the origin, be OA where O(0,0,0) and the point A(2,1,1) and PQ be the line joining the points P(3,5,-1) and Q(4,3,-1) .

    Then the direction ratios of the line OA will be (2-0),(1-0),\ and\ (1-0) = 2,1,1 and that of line PQ will be

    (4-3),(3-5),\ and\ (-1+1) = 1,-2,0

    So to check whether line OA is perpendicular to line PQ then,

    Applying the relation we know,

    a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    \Rightarrow 2(1)+1(-2)+1(0) = 2-2+0 = 0

    Therefore OA is perpendicular to line PQ.

    Question:2 If l 1 , m 1 , n 1 and l 2 , m 2 , n 2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m_{1}n_{2}-m_{2}n_{1}, n_{1}l_{2}-n_{2}l_{1},l_{1}m_{2}-l_{2}m_{1} .

    Answer:

    Given that l_1,m_1,n_1\ and\ l_2,m_2,n_2 are the direction cosines of two mutually perpendicular lines.

    Therefore, we have the relation:

    l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2} = 0 .........................(1)

    l_{1}^2+m_{1}^2+n_{1}^2 =1\ and\ l_{2}^2+m_{2}^2+n_{2}^2 =1 .............(2)

    Now, let us assume l,m,n be the new direction cosines of the lines which are perpendicular to the line with direction cosines. l_1,m_1,n_1\ and\ l_2,m_2,n_2

    Therefore we have, ll_{1}+mm_{1}+nn_{1} = 0 \and\ ll_{2}+mm_{2}+nn_{2} = 0

    Or, \frac{l}{m_{1}n_{2}-m_{2}n_{1}} = \frac{m}{n_{1}l_{2}-n_{2}l_{1}} = \frac{n}{l_{1}m_{2}-l_{2}m_{1}}

    \Rightarrow \frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2}

    \Rightarrow \frac{l^2+m^2+n^2}{(m_{1}n_{2}-m_{2}n_{1})^2 +(n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2} ......(3)

    So, l,m,n are the direction cosines of the line.

    where, l^2+m^2+n^2 =1 ........................(4)

    Then we know that,

    \Rightarrow (l_{1}^2+m_{1}^2+n_{1}^2)(l_{2}^2+m_{2}^2+n_{2}^2) - (l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2})^2

    = (m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2

    So, from the equation (1) and (2) we have,

    (1)(1) -(0) =(m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2

    Therefore, (m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2 =1 ..(5)

    Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

    \frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2} =1

    Therefore we have the direction cosines of the required line as;

    l =m_{1}n_{2} - m_{2}n_{1}

    m =n_{1}l_{2} - n_{2}l_{1}

    n =l_{1}m_{2} - l_{2}m_{1}

    Question:3 Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

    Answer:

    Given direction ratios a,b,c and b-c,\ c-a,\ a-b .

    Thus the angle between the lines A is given by;

    A = \left | \frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}.\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \right |

    \Rightarrow \cos A = 0

    \Rightarrow A = \cos^{-1}(0) = 90^{\circ} a

    Thus, the angle between the lines is 90^{\circ}

    Question:4 Find the equation of a line parallel to x-axis and passing through the origin.

    Answer:

    Equation of a line parallel to the x-axis and passing through the origin (0,0,0) is itself x-axis .

    So, let A be a point on the x-axis.

    Therefore, the coordinates of A are given by (a,0,0) , where a\epsilon R .

    Now, the direction ratios of OA are (a-0) =a,0 , 0

    So, the equation of OA is given by,

    \frac{x-0}{a} = \frac{y-0}{0} = \frac{z-0}{0}

    or \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a

    Thus, the equation of the line parallel to the x-axis and passing through origin is

    \frac{x}{1} = \frac{y}{0} = \frac{z}{0}

    Question:5 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

    Answer:

    Direction ratios of AB are (4-1),(5-2),(7-3) = 3,3,4

    and Direction ratios of CD are (2-(-4)), (9-3), (2-(-6)) = 6,6,8

    So, it can be noticed that, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} = \frac{1}{2}

    Therefore, AB is parallel to CD.

    Thus, we can easily say the angle between AB and CD which is either 0^{\circ}\ or\ 180^{\circ} .

    Question:6 If the lines \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2} and \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5} are perpendicular, find the value of k.

    Answer:

    Given both lines are perpendicular so we have the relation; a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

    For the two lines whose direction ratios are known,

    a_{1},b_{1},c_{1}\ and\ a_{2},b_{2},c_{2}

    We have the direction ratios of the lines, \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2} and \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5} are -3,2k,2 and 3k,1,-5 respectively.

    Therefore applying the formula,

    -3(3k)+2k(1)+2(-5) = 0

    \Rightarrow -9k +2k -10 = 0

    \Rightarrow7k=-10 or k= \frac{-10}{7}

    \therefore For, k= \frac{-10}{7} the lines are perpendicular.

    Question:7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0

    Answer:

    Given that the plane is passing through the point A (1,2,3) so, the position vector of the point A is \vec{r_{A}} = \widehat{i}+2\widehat{j}+3\widehat{k} and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0 whose direction ratios are 1,2,\ and\ -5 and the normal vector is \vec{n} = \widehat{i}+2\widehat{j}-5\widehat{k}

    So, the equation of a line passing through a point and perpendicular to the given plane is given by,

    \vec{l} = \vec{r} + \lambda\vec{n} , where \lambda \epsilon R

    \Rightarrow \vec{l} = (\widehat{i}+2\widehat{j}+3\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}-5\widehat{k})

    Question:8 Find the equation of the plane passing through (a, b, c) and parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 .

    Answer:

    Given that the plane is passing through (a,b,c) and is parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2

    So, we have

    The position vector of the point A(a,b,c) is, \vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}

    and any plane which is parallel to the plane, \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 is of the form,

    \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\lambda . .......................(1)

    Therefore the equation we get,

    ( a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=\lambda

    Or, a+b+c = \lambda

    So, now substituting the value of \lambda = a+b+c in equation (1), we get

    \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c .................(2)

    So, this is the required equation of the plane .

    Now, substituting \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k} in equation (2), we get

    (x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c

    Or, x+y+z = a+b+c

    Question:9 Find the shortest distance between lines \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k}) and \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}) .

    Answer:

    Given lines are;

    \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k}) and

    \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})

    So, we can find the shortest distance between two lines \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} and \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} by the formula,

    d = \left | \frac{(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}}\times\vec{b_{2}} \right |} \right | ...........................(1)

    Now, we have from the comparisons of the given equations of lines.

    \vec{a_{1}} = 6\widehat{i}+2\widehat{j}+2\widehat{k} \vec{b_{1}} = \widehat{i}-2\widehat{j}+2\widehat{k}

    \vec{a_{2}} = -4\widehat{i}-\widehat{k} \vec{b_{2}} = 3\widehat{i}-2\widehat{j}-2\widehat{k}

    So, \vec{a_{2}} -\vec{a_{1}} = (-4\widehat{i}-\widehat{k}) -(6\widehat{i}+2\widehat{j}+2\widehat{k}) = -10\widehat{i}-2\widehat{j}-3\widehat{k}

    and \Rightarrow \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &-2 &2 \\ 3& -2 &-2 \end{vmatrix} = (4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}

    =8\widehat{i}+8\widehat{j}+4\widehat{k}

    \therefore \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{8^2+8^2+4^2} =12

    (\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}}) = (8\widehat{i}+8\widehat{j}+4\widehat{k}).(-10\widehat{i}-2\widehat{j}-3\widehat{k}) = -80-16-12 =-108 Now, substituting all values in equation (3) we get,

    d = | \frac{-108}{12}| = 9

    Hence the shortest distance between the two given lines is 9 units.

    Question:10 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

    Answer:

    We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) is given by the relation;

    \frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

    and the line passing through the points, \frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}

    \implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)

    \implies x=5-2k,\ y=3k+1,\ z=6-5k

    And any point on the line is of the form (5-2k,3k+1,6-5k) .

    So, the equation of the YZ plane is x=0

    Since the line passes through YZ- plane,

    we have then,

    5-2k = 0

    \Rightarrow k =\frac{5}{2}

    or 3k+1 = 3(\frac{5}{2})+1 = \frac{17}{2} and 6-5k= 6-5(\frac{5}{2}) = \frac{-13}{2}

    So, therefore the required point is \left ( 0,\frac{17}{2},\frac{-13}{2} \right )

    Question : 11 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

    Answer:

    We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) is given by the relation;

    \frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

    and the line passing through the points, \frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}

    \implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)

    \implies x=5-2k,\ y=3k+1,\ z=6-5k

    And any point on the line is of the form (5-2k,3k+1,6-5k) .

    So, the equation of ZX plane is y=0

    Since the line passes through YZ- plane,

    we have then,

    3k+1 = 0

    \Rightarrow k =-\frac{1}{3}

    or 5-2k = 5-2\left ( -\frac{1}{3} \right ) = \frac{17}{3} and 6-5k= 6-5(\frac{-1}{3}) = \frac{23}{3}

    So, therefore the required point is \left ( \frac{17}{3},0,\frac{23}{3} \right )

    Question:12 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

    Answer:

    We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) is given by the relation;

    \frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

    and the line passing through the points, (3,-4,-5)\ and\ (2,-3,1) .

    \implies \frac{x-3}{2-3} = \frac{y+4}{-3+4} = \frac{z+5}{1+5} = k\ (say)

    \implies \frac{x-3}{-1} = \frac{y+4}{-1} = \frac{z+5}{6} = k\ (say)

    \implies x=3-k,\ y=k-4,\ z=6k-5

    And any point on the line is of the form. (3-k,k-4,6k-5)

    This point lies on the plane, 2x+y+z = 7

    \therefore 2(3-k)+(k-4)+(6k-5) = 7

    \implies 5k-3=7

    or k =2 .

    Hence, the coordinates of the required point are (3-2,2-4,6(2)-5) or (1,-2,7) .

    Question:13 Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

    Answer:

    Given

    two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

    the normal vectors of these plane are

    n_1=\hat i+2\hat j+ 3\hat k

    n_2=3\hat i+3\hat j+ \hat k

    Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

    \vec n = \vec n_1\times\vec n_2

    \vec n = (\hat i+2\hat j +3\hat k )\times (3\hat i + 3\hat j +\hat k)

    \vec n =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ 3& 3& 1 \end{vmatrix}=\hat i(2-9)-\hat j(1-9)+\hat k (3-6)

    \vec n =-7\hat i+8\hat j-3\hat k

    Now, as we know

    the equation of a plane in vector form is :

    \vec r\cdot\vec n=d

    \vec r\cdot(-7\hat i+8\hat j-3\hat k)=d

    Now Since this plane passes through the point (-1,3,2)

    (-\hat i+3\hat j+2\hat k)\cdot(-7\hat i+8\hat j-3\hat k)=d

    7+24-6=d

    d=25

    Hence the equation of the plane is

    \vec r\cdot(-7\hat i+8\hat j-3\hat k)=25

    Question:14 If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0 then find the value of p.

    Answer:

    Given that the points A(1,1,p) and B(-3,0,1) are equidistant from the plane

    \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0

    So we can write the position vector through the point (1,1,p) is \vec{a_{1}} = \widehat{i}+\widehat{j}+p\widehat{k}

    Similarly, the position vector through the point B(-3,0,1) is

    \vec{a_{2}} = -4\widehat{i}+\widehat{k}

    The equation of the given plane is \overrightarrow{r}.(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 =0

    and We know that the perpendicular distance between a point whose position vector is \vec{a} and the plane, \vec{n} = 3\widehat{i}+4\widehat{j}-12\widehat{k} and d =-13

    Therefore, the distance between the point A(1,1,p) and the given plane is

    D_{1} = \frac{\left | (\widehat{i}+\widehat{j}+p\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}

    D_{1} = \frac{\left | 3+4-12p+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}

    D_{1} = \frac{\left | 20-12p \right |}{13} nbsp; .........................(1)

    Similarly, the distance between the point B(-1,0,1) , and the given plane is

    D_{2} = \frac{\left | (-3\widehat{i}+\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}

    D_{2} = \frac{\left |-9-12+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}

    D_{2} = \frac{8}{13} .........................(2)

    And it is given that the distance between the required plane and the points, A(1,1,p) and B(-3,0,1) is equal.

    \therefore D_{1} =D_{2}

    \implies \frac{\left | 20-12p \right |}{13} =\frac{8}{13}

    therefore we have,

    \implies 12p =12

    or p =1 or p = \frac{7}{3}

    Question:15 Find the equation of the plane passing through the line of intersection of the planes \overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1 and \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0 and parallel to x-axis.

    Answer:

    So, the given planes are:

    \overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1 and \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0

    The equation of any plane passing through the line of intersection of these planes is

    [\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )-1] + \lambda \left [ \overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4\right ] = 0

    \vec{r}.[(2\lambda+1)\widehat{i}+(3\lambda+1)\widehat{j}+(1-\lambda)\widehat{k}]+(4\lambda+1) = 0 ..............(1)

    Its direction ratios are (2\lambda+1) , (3\lambda+1), and (1-\lambda) = 0

    The required plane is parallel to the x-axis.

    Therefore, its normal is perpendicular to the x-axis.

    The direction ratios of the x-axis are 1,0, and 0.

    \therefore \1.(2\lambda+1) + 0(\3\lambda+1)+0(1-\lambda) = 0

    \implies 2\lambda+1 = 0

    \implies \lambda = -\frac{1}{2}

    Substituting \lambda = -\frac{1}{2} in equation (1), we obtain

    \implies \vec{r}.\left [ -\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k} \right ]+(-3)= 0

    \implies \vec{r}(\widehat{j}-3\widehat{k})+6= 0

    So, the Cartesian equation is y -3z+6 = 0

    Question:16 If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

    Answer:

    We have the coordinates of the points O(0,0,0) and P(1,2,-3) respectively.

    Therefore, the direction ratios of OP are (1-0) = 1, (2-0)=2,\ and\ (-3-0)=-3

    And we know that the equation of the plane passing through the point (x_{1},y_{1},z_{1}) is

    a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0 where a,b,c are the direction ratios of normal.

    Here, the direction ratios of normal are 1,2, and -3 and the point P is (1,2,-3) .

    Thus, the equation of the required plane is

    1(x-1)+2(y-2)-3(z+3) = 0

    \implies x+2y -3z-14 = 0

    Question:17 Find the equation of the plane which contains the line of intersection of the planes \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0 and which is perpendicular to the plane \overrightarrow{r}.(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0

    Answer:

    The equation of the plane passing through the line of intersection of the given plane in \overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0

    \left [ \vec{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4 \right ]+\lambda\left [ \vec{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5 \right ] = 0

    \vec{r}.[(2\lambda+1)\widehat{i}+(\lambda+2)\widehat{j}+(3-\lambda)\widehat{k}]+(5\lambda-4)= 0 ,,,,,,,,,,,,,(1)

    The plane in equation (1) is perpendicular to the plane, 1633928330199 Therefore 5(2\lambda+1)+3(\lambda+2) -6(3-\lambda) = 0

    \implies 19\lambda -7 = 0

    \implies \lambda = \frac{7}{19}

    Substituting \lambda = \frac{7}{19} in equation (1), we obtain

    \implies \vec{r}.\left [ \frac{33}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k} \right ] -\frac{41}{19} = 0

    \implies \vec{r}.(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0 .......................(4)

    So, this is the vector equation of the required plane.

    The Cartesian equation of this plane can be obtained by substituting \implies \vec{r}= (x\widehat{i}+y\widehat{j}+z\widehat{k}) in equation (1).

    (x\widehat{i}+y\widehat{j}+z\widehat{k}).(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0

    Therefore we get the answer 33x+45y+50z -41 = 0

    Question:18 Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) and the plane \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5 .

    Answer:

    Given,

    Equation of a line :

    \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )

    Equation of the plane

    \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5

    Let's first find out the point of intersection of line and plane.

    putting the value of \vec r into the equation of a plane from the equation from line

    \left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5

    (2+3\lambda)-(4\lambda -1)+(2+2\lambda)=5

    \lambda+5=5

    \lambda=0

    Now, from the equation, any point p in line is

    P=(2+3\lambda,4\lambda-1,2+2\lambda)

    So the point of intersection is

    P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)

    SO, Now,

    The distance between the points (-1,-5,-10) and (2,-1,2) is

    d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}

    d=\sqrt{169}=13

    Hence the required distance is 13.

    Question:19 Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5 and \overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6 .

    Answer:

    Given

    A point through which line passes

    \vec a=\hat i+2\hat j+3\hat k

    two plane

    \overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5 And

    \overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6

    it can be seen that normals of the planes are

    \vec n_1=\hat i-\hat j+2\hat k

    \vec n_2=3\hat i+\hat j+\hat k
    since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

    So, a vector perpendicular to both these normal vector is

    \vec d=\vec n_1\times\vec n_2

    \vec d=\begin{vmatrix} \hat i &\hat j & \hat k\\ 1 &-1 &2 \\ 3& 1 & 1 \end{vmatrix}=\hat i(-1-2)-\hat j(1-6)+\hat k(1+3)

    \vec d=-3\hat i+5\hat j+4\hat k

    Now a line which passes through \vec a and parallels to \vec d is

    L=\vec a+\lambda\vec d

    So the required line is

    L=\vec a+\lambda\vec d

    L=\hat i+2\hat j+3\hat k+\lambda(-3\hat i+5\hat j+4\hat k)

    L=(1-3\lambda)\hat i+(2+5\lambda)\hat j+(3+4\lambda)\hat k

    Questio n: 20 Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

    \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}

    Answer:

    Given

    Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)

    Now the two vectors which are parallel to the two lines are

    \vec a= 3\hat i-16\hat j+7\hat k and

    \vec b= 3\hat i+8\hat j-5\hat k

    As we know, a vector perpendicular to both vectors \vec a and \vec b is \vec a\times\vec b , so

    \vec a\times\vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3& -16 &7 \\ 3&8 &-5 \end{vmatrix}=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)

    \vec a\times\vec b=24\hat i+36\hat j+72\hat k

    A vector parallel to this vector is

    \vec d=2\hat i+3\hat j+6\hat k

    Now as we know the vector equation of the line which passes through point p and parallel to vector d is

    L=\vec p+\lambda \vec d

    Here in our question, give point p = (1,2,-4) which means position vector of this point is

    \vec p = \hat i +2\hat j-4\hat k

    So, the required line is

    L=\vec p+\lambda \vec d

    L=\hat i+2\hat j-4\hat k +\lambda (2\hat i+3\hat j+6\hat k)

    L=(2\lambda +1)\hat i+(2+3\lambda)\hat j+(6\lambda-4)\hat k

    Question:21 Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}} .

    Answer:

    The equation of plane having a, b and c intercepts with x, y and z-axis respectively is given by
    \frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 1
    The distance p of the plane from the origin is given by
    \\p = \left | \frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ p = \left | \frac{-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ \frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
    Hence proved

    Question:22 Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

    (A) 2 units (B) 4 units (C) 8 units (D) \frac{2}{\sqrt{29}}unit

    Answer:

    Given equations are
    2x+3y+4z= 4 \ \ \ \ \ \ \ \ \ -(i)
    and
    4x+6y+8z= 12\\ 2(2x+3y+4z)= 12\\ 2x+3y+4z = 6 \ \ \ \ \ \ \ \ \ \ -(ii)
    Now, it is clear from equation (i) and (ii) that given planes are parallel
    We know that the distance between two parallel planes ax+by +cz = d_1 \ and \ ax+by +cz = d_2 is given by
    D= \left | \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}} \right |
    Put the values in this equation
    we will get,
    D= \left | \frac{6-4}{\sqrt{2^2+3^2+4^2}} \right |
    D= \left | \frac{2}{\sqrt{4+9+16}} \right |= \left | \frac{2}{\sqrt{29}} \right |
    Therefore, the correct answer is (D)

    Question:23 The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

    (A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through \left ( 0,0,\frac{5}{4} \right )

    Answer:

    Given equations of planes are
    2x-y+4z=5 \ \ \ \ \ \ \ \ \ -(i)
    and
    5x-2.5y+10z=6\\ 2.5(2x-y+4z)=6\\ 2x-y+4z= 2.4 \ \ \ \ \ \ \ \ \ -(ii)
    Now, from equation (i) and (ii) it is clear that given planes are parallel to each other
    \because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{2}{2}= \frac{-1}{-1}=\frac{4}{4}\Rightarrow 1=1=1
    Therefore, the correct answer is (B)

    If you are looking for exercises solutions for chapter 3d Geometry class 12 then they are listed below.

    More about NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

    • All the important topics are covered in the maths chapter 11 class 12 NCERT solutions.

    • A total of 36 questions in 3 exercises are given in this Maths chapter 11 class 12 solutions.

    • All these NCERT questions are solved and explained in the NCERT solutions for class 12 maths chapter 11 three dimensional geometry article to clear your doubts.

    • In this NCERT Solutions for Class 12 Maths Chapter 11 PDF Download, we deal with formulas like-

    • If l, m, n are the direction cosines of a line, thenl^2+m^2+n^2=1.

    • Q(x_{2}, y_{2}, z_{2})andP(x_{1}, y_{1}, z_{1})Direction cosines of a line joining two pointsPQ=\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2+(Z_2-Z_1)^2}are\frac{X_2-X_1}{PQ},\:\frac{Y_2-Y_1}{PQ},\:\frac{Z_2-Z_1}{PQ}, where

    • If l, m, n are the direction cosines and a, b, c are the direction of a line then-

    • \dpi{80} l=\frac{a}{\sqrt{a^2+b^2+c^2}} ; m=\frac{b}{\sqrt{a^2+b^2+c^2}}; n=\frac{c}{\sqrt{a^2+b^2+c^2}}

    Also read,

    Three Dimensional Geometry Class 12 - Topics

    11.1 Introduction

    11.2 Direction Cosines and Direction Ratios of a Line

    11.2.1 Relation between the direction cosines of a line

    11.2.2 Direction cosines of a line passing through two points

    11.3 Equation of a Line in Space

    11.3.1Equation of a line through a given point and parallel to a given vector b

    11.3.2 Equation of a line passing through two given points

    11.4 Angle between Two Lines

    11.5 Shortest Distance between Two Lines

    11.5.1 Distance between two skew lines

    11.5.2 Distance between parallel lines

    11.6 Plane

    11.6.1 Equation of a plane in normal form

    11.6.2 Equation of a plane perpendicular to a given vector and passing through a given point

    11.6.3 Equation of a plane passing through three noncollinear points

    11.6.4 Intercept form of the equation of a plane

    11.6.5 Plane passing through the intersection of two given planes

    11.7 Coplanarity of Two Lines

    11.8 Angle between Two Planes

    11.9 Distance of a Point from a Plane

    11.10 Angle between a Line and a Plane

    NCERT solutions for class 12 maths - Chapter wise

    Key Features of Three Dimensional Geometry Class 12 Solutions

    Chapter 11 of Class 12 Maths is titled "Three Dimensional Geometry." NCERT Solutions for this chapter provide step-by-step solutions to all the exercises and problems included in the textbook. Some of the key features of NCERT Solutions for Class 12 Maths Chapter 11 are:

    1. Comprehensive Coverage: The ch 11 maths class 12 solutions cover all the important topics and concepts discussed in the chapter, including the coordinate axes and coordinate planes in three dimensions, distance between two points, section formula, direction cosines and direction ratios of a line, angle between two lines, equation of a line and a plane, and the distance of a point from a plane.

    2. Simple and Clear Explanation: The class 12 maths ch 11 question answer are written in a simple and clear language that makes it easy for students to understand even the most complex concepts.

    3. Step-by-step Approach: The class 12 maths ch 11 question answer are provided in a step-by-step manner, making it easy for students to follow and learn.

    NCERT solutions for class 12 subject wise

    NCERT Solutions Class Wise

    Benefits of three dimensional geometry class 12 NCERT solutions

    • Class 12 Maths Chapter 11 NCERT solutions are very easy for you to understand the concepts as they are explained in a step-by-step manner.

    • NCERT Class 12 Maths solutions chapter 11 will give you some new insight into the concepts.

    • Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 11 three dimensional geometry as these questions are answered by the experts who know how best to answer the questions in the board exam.

    • You should solve the miscellaneous exercise also, to develop a grip on the concepts. Here, you will get solutions for miscellaneous exercise too.

    • Three dimensional geometry class 12 ncert solutions PDF Download will also be made available soon.

    NCERT Books and NCERT Syllabus

    Frequently Asked Question (FAQs)

    1. What is the weightage of the chapter three-dimensional geometry for the CBSE board exam?

    The concepts of vector algebra and Three Dimensional Geometry can be used interchangeably. if two chapters are combined, vector algebra & three-dimensional geometry has a 17% weightage in the 12th board maths final exam. after getting command of these concepts it becomes easy for students to score well in the exam therefore NCERT Notes, NCERT syllabus, and NCERT textbooks are recommended.

    2. Would it be accurate to say that three dimensional geometry class 12 ncert solutions is the most helpful study material for students during revision?

    Undoubtedly, NCERT Solutions for Class 12 Maths Chapter 11 stand out as the top study material aiding students in effortlessly revising complex concepts. The solutions present a well-reasoned explanation to facilitate student learning. The team of experts at Careers360 has carefully crafted step-by-step solutions that encourage students to employ an analytical thinking approach. Moreover, these solutions can be cross-referenced to gain insights into alternative methods to solve textbook problems.

    3. How the NCERT solutions are helpful in the CBSE board exam?

    Only knowing the answer is not enough to score good marks in the exam. One should know how best to answer in order to get good marks. NCERT solutions are provided by the experts who know how best to write answer in the board exam in order to get good marks.

    4. What are the important topics covered in the chapter of ncert solutions class 12 maths chapter 11?

     Class 12 chapter 11 maths Maths covers the following important topics:

    • Introduction (11.1)

    • Direction cosines and direction ratios of a line (11.2)

    • Equation of a line in space (11.3)

    • Angle between two lines (11.4)

    • Shortest distance between two lines (11.5)

    • Plane (11.6)

    • Coplanarity of two lines (11.7)

    • Angle between two planes (11.8)

    • Distance of a point from a plane (11.9)

    • Angle between a line and a plane (11.10)

    5. How class 12 math chapter 11 miscellaneous exercises are useful?

    The ncert solutions class 12 maths chapter 11 miscellaneous is useful for students in several ways. These exercises include additional problems that are not part of the main textbook exercises but are relevant to the chapter's concepts.

    Firstly, NCERT solutions for class 12 maths chapter 11 miscellaneous exercise help students to gain a deeper understanding of the concepts covered in Chapter 11. 

    Secondly, these ch 11 miscellaneous class 12 exercises help students to test their problem-solving abilities and improve their skills. 

    Thirdly, solving the miscellaneous exercise problems for 3d class 12 ncert solutions can help students to prepare for their exams.

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    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    Orthotist and Prosthetist

    Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

    6 Jobs Available
    Veterinary Doctor

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    5 Jobs Available
    Pathologist

    A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

    5 Jobs Available
    Speech Therapist
    4 Jobs Available
    Gynaecologist

    Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

    4 Jobs Available
    Oncologist

    An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

    3 Jobs Available
    Audiologist

    The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

    3 Jobs Available
    Dental Surgeon

    A Dental Surgeon is a professional who possesses specialisation in advanced dental procedures and aesthetics. Dental surgeon duties and responsibilities may include fitting dental prosthetics such as crowns, caps, bridges, veneers, dentures and implants following apicoectomy and other surgical procedures.

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

    Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    Multimedia Specialist

    A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

    2 Jobs Available
    Producer

    An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

    They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

    Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

    3 Jobs Available
    Travel Journalist

    The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    SEO Analyst

    An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns. 

    2 Jobs Available
    Welding Engineer

    Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

    5 Jobs Available
    QA Manager
    4 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager
    3 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    AWS Solution Architect

    An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

    4 Jobs Available
    QA Manager
    4 Jobs Available
    Azure Administrator

    An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

    4 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager
    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
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