NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

Edited By Komal Miglani | Updated on May 12, 2025 03:31 PM IST

Three-dimensional geometry bridges imagination and mathematics, giving form to the invisible and precision to the abstract. Three-dimensional geometry is an important topic of Class 12 Maths, which helps understand the geometrical relationship between points, lines, and planes in 3-D space. In this chapter, we study basic geometric concepts and learn about new methods to solve problems in real-life scenarios like finding the coordinates of an aeroplane, satellite, etc. Class 12 NCERT solutions OF Three-dimensional geometry cover all topics, including the distance formula in 3D, direction cosines and direction ratios, equations of lines and planes, and many more. We will also learn to calculate the shortest distance between a point and a plane using the necessary formulae and detailed step-by-step solutions.

This Story also Contains
  1. NCERT Solution for Class 12 Maths Chapter 11 Solutions: Download PDF
  2. NCERT Class 12 Maths Chapter 11 Three-Dimensional Geometry - Important Formulae
  3. NCERT Solutions for Class 12 Maths Chapter 11: Exercise Questions
  4. Class 12 Maths NCERT Chapter 11: Extra Question
  5. Approach to Solve Questions of Three-Dimensional Geometry Class 12
  6. What Extra Should Students Study Beyond NCERT for JEE?
  7. NCERT Solutions for Class 12 Maths: Chapter Wise
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

In three dimensions, every angle is a perspective, and every shape tells a story in space. NCERT Solutions for Class 12 Maths Chapter 11 Three-Dimensional Geometry, prepared by experts at Careers360, offers detailed informational study material for students preparing for the CBSE Class 12 board exam according to the latest CBSE Syllabus. For syllabus, notes, and PDF, refer to this link: NCERT.

NCERT Solution for Class 12 Maths Chapter 11 Solutions: Download PDF

Students who wish to access the Class 12 Maths Chapter 11 NCERT Solutions can click on the link below to download the complete solution in PDF.

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NCERT Class 12 Maths Chapter 11 Three-Dimensional Geometry - Important Formulae

Distance Formula:

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

$A B=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

The distance between a point A(x, y, z) and the origin O(0, 0, 0) is given by:

$\mathrm{OA}=\sqrt{ }\left(\mathrm{x}^2+\mathrm{y}^2+z^2\right)$

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: $\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)$

External Division: $\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)$

Midpoint Formula: The coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are:

$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

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

$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$

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

$\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right)$

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

$\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$

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:

$l^2+m^2+n^2=1$

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

$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}$

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

Two lines are parallel if: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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:

$\left|l\left(x_2-x_1\right)+m\left(y_2-y_1\right)+n\left(z_2-z_1\right)\right|$

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

  • General form: $a x+b y+c z+d=0$ (where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are not all zero)
  • Normal form:$\mathrm{lx}+\mathrm{my}+\mathrm{nz}=\mathrm{p}$
  • Plane through a point $\left(x_1, y_1, z_1\right)$ : $a(x-$$ \left.x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$
  • Intercept form: $(\mathrm{x} / \mathrm{a})+(\mathrm{y} / \mathrm{b})+(\mathrm{z} / \mathrm{c})=1$
  • Vector form: $(r-a). n=0$ or $r . n=a . n$

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

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Angle between Two Lines: $\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|$=$\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}\right|$

Shortest distance between skew lines:$\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is

$
\left|\frac{\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|
$

Cartesian form:

The shortest distance between the lines

$\begin{aligned} & l_1: \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \\ & l_2: \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\end{aligned}$

is

$\left|\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}\right|$

Distance between parallel lines: $\vec{r}=\vec{a}_1+\lambda \vec{b}$ and $\vec{r}=\vec{a}_2+\mu \vec{b}$ is

$
\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|
$

If $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ are the equations of two lines, then the acute angle between the two lines is given by $\cos \theta=\left|l_1 l_2+m_1 m_2+n_1 n_2\right|$.

NCERT Solutions for Class 12 Maths Chapter 11: Exercise Questions

Class 12 Maths chapter 11 solutions - Exercise: 11.1
Page number: 381
Total questions: 5

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 an equal angle with the coordinate axes. Then,

Let the common angle made be $\alpha$ with each coordinate axis.

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;

A line having direction ratio -18 has direction cosine:

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

A line having direction ratio 12 has direction cosine:

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

A 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 are -2 times.

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

Hence, 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 the 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;

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

Class 12 Maths chapter 11 solutions - Exercise: 11.2
Page number: 389-390
Total questions: 15

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

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

$\Rightarrow$$ \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 $L_{2}\ and\ L_{3}$ ;

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

$\Rightarrow$$ \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 $L_{3}\ and\ L_{1}$ ;

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

$\Rightarrow$$ \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 that 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 the 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 $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =0$

$\Rightarrow$$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 )$

$\Rightarrow$$ 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 them;

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

Question 5: Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\widehat{i}-\widehat{j}+4\widehat{k}$ and is in the direction $\widehat{i}+2\widehat{j}-\widehat{k}$.

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

(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:

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

(ii) 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 9: Find the angle between the following pair of lines:

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

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

Answer:

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

(ii) 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 have 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 10: 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 11: 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 12: 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 13: 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 the determinant

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

$\Rightarrow$$ 4(-6+2)-6(7-1)+8(-14+6)$

$\Rightarrow$$ -16-36-64$

$\Rightarrow$$-116$

Now, calculating the denominator,

$\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2}$

$\Rightarrow$$ \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 14: 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 15: 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 is the shortest distance between the two given lines.

Class 12 Maths chapter 11 solutions - Miscellaneous Exercise
Page number: 390-391
Total questions: 5

Question 1: 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;

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

$\Rightarrow \cos A = 0$

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

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

Question 2: Find the equation of a line parallel to the 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 the 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 the origin is

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

Question 3: 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 4: 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 5: 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 are (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 is parallel to vector d is

$L=\vec p+\lambda \vec d$

Here in our question, give point p = (1,2,-4), which means the position vector of this point is

$\vec p = \hat i +2\hat j-4\hat k$

So, the required line is

$L=\vec p+\lambda \vec d$

$\Rightarrow$$L=\hat i+2\hat j-4\hat k +\lambda (2\hat i+3\hat j+6\hat k)$

$\Rightarrow$$L=(2\lambda +1)\hat i+(2+3\lambda)\hat j+(6\lambda-4)\hat k$

Also read,

Class 12 Maths NCERT Chapter 11: Extra Question

Question: Find the distance between the parallel planes $2 x+y+2 z=8$ and $4 x+2 y+4 z+5=0$

Solution:
Given Planes:
$
\begin{aligned}
& 2 x+y+2 z=8 \\
& \Rightarrow 4 x+2 y+4 z=16 \\
& 4 x+2 y+4 z=-5
\end{aligned}
$
Distance between two parallel planes:

$
\begin{aligned}
& d=\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}} \\
& d=\frac{|16-(-5)|}{\sqrt{4^2+2^2+4^2}} \\
& d=\frac{21}{\sqrt{36}}=\frac{21}{6} \\
& d=\frac{7}{2} \text { units }
\end{aligned}
$
Hence, the correct answer is $\frac72$ units.

Approach to Solve Questions of Three-Dimensional Geometry Class 12

Strong foundation of basic concepts: Get yourself familiar with direction cosines (l, m, n) and direction ratios (a,b,c) to answer all the three-dimensional line questions. Also, comfortable switching the equation of a line between vector and Cartesian form.

Cosine rule identity: If $l, m, n$ are the direction cosines of a line, then $l^2+m^2+n^2=1$. If this formula doesn't hold, then check for direction ratios.

Avoid confusion: Do not mix up the position vector with the direction vector. Read the question carefully and answer accordingly.

Compare the lines: Check whether two lines are parallel, intersecting, or skew using direction ratios and point substitution.

Memorise formulas: Three-dimensional geometry has many important formulas which are necessary to solve the problems. We have provided important formulas at the beginning of the article. Memorise them from time to time.

What Extra Should Students Study Beyond NCERT for JEE?

NCERT Solutions for Class 12 Maths: Chapter Wise

Also read,

NCERT solutions for class 12 subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Solutions Class Wise

Given below are the class-wise solutions of class 12 NCERT:

NCERT Books and NCERT Syllabus

Here are some useful links for the NCERT books and the NCERT syllabus for class 12.

Frequently Asked Questions (FAQs)

1. How to find the angle between two lines in three-dimensional geometry?

To find the angle between two lines in three-dimensional space, we use the dot product formula. The formula is:

cos(θ) = (a · b) / (|a| |b|)

where a and b are the direction vectors of the two lines. This formula helps determine the angle between two lines by relating the vectors' orientation in space, and the result will give the angle between the lines.

2. What is the formula for the distance between two parallel lines in 3D geometry?

The distance between two parallel lines in 3D space can be calculated by finding the perpendicular distance between a point on one line and the other line. If we have two parallel lines with direction vector d, and the lines pass through points P1 and P2, then the formula for the distance D is:

D = |(P2 - P1) × d| / |d|
Here, P2-P1 is the vector between any two points on the lines, and the cross product gives the area of the parallelogram formed by the two vectors. Dividing by the magnitude of the direction vector d yields the shortest distance between the lines.

3. How to calculate the equation of a plane passing through three points?

To find the equation of a plane passing through three given points in 3D, let the three points be A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3). First, calculate two vectors lying in the plane, say AB = (x2 - x1, y2 - y1, z2 - z1) and AC = (x3 - x1, y3 - y1, z3 - z1). Then, find the cross product of these vectors to get the normal vector n to the plane. The equation of the plane is then given by:

n · (r - r?) = 0

where r is a point on the plane, (r - r?) is a known point (such as A ), and n is the normal vector obtained from the cross-product.

4. What is the shortest distance between skew lines in 3D space?

The shortest distance between skew lines (lines that are not parallel and do not intersect) is the perpendicular distance between them. To find it, we first find two points, one on each line, say P1 on line 1 and P2 on line 2 . Next, we compute the vector between these two points, P1P2. The shortest distance is given by the formula:

D = |(P1 P2) · (d1 × d2)| / |d1 × d2|

where d1 and d2 are the direction vectors of the two lines, and d1 × d2 gives a vector perpendicular to both lines. This distance represents the shortest path between the two skew lines.

5. What is the Cartesian equation of a line in 3D geometry?

The Cartesian equation of a line in 3D geometry is given by the system of equations:

(x - x0) / a = (y - y0) / b = (z - z0) / c
Here, (x0, y0, z0) is a point on the line, and (a, b, c) are the direction ratios of the line. This equation relates the coordinates of any point on the line.

Articles

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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