NCERT Solutions for Exercise 11.2 Class 12 Maths Chapter 11- Three Dimensional Geometry

NCERT Solutions for Exercise 11.2 Class 12 Maths Chapter 11- Three Dimensional Geometry

Edited By Ramraj Saini | Updated on Dec 04, 2023 09:10 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.2

NCERT Solutions for Exercise 11.2 Class 12 Maths Chapter 11 Three Dimensional Geometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 11.2 Class 12 Maths chapter 11 is about the lines in space and the equation of these lines in cartesian and vector form. NCERT solutions for Class 12 Maths chapter 11 exercise 11.2 also covers the topic of the distance between lines. The topics covered in the exercise 11.2 Class 12 Maths are very important if the CBSE Class 12 Maths Previous Paper is considered. The main focus of solving Class 12 Maths chapter 11 exercise 11.2 should be to check whether the concepts are grasped or not.

12th class Maths exercise 11.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Three Dimensional Geometry Class 12th Chapter 11 -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 :

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.

More About NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Ten solved questions are given prior to the exercise 11.2 Class 12 Maths. And 17 questions are covered in Class 12 Maths chapter 11 exercise 11.2. Broadly speaking Class 12th Maths chapter 11 exercise 11.2 covers questions related to the equation of a line parallel to a given vector and that passes through a given point, the line passing through two given points, the angle between lines, the smallest distance between two lines and distance between the skew lines and parallel lines.

Also Read| Three Dimensional Geometry Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

  • Solving the given NCERT examples for topic 11.3 and exercise 11.2 Class 12 Maths help to score well in the exam
  • Question of similar type in Class 12th Maths chapter 11 exercise 11.2 can be expected for CBSE board exam.
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Key Features Of NCERT Solutions for Exercise 11.2 Class 12 Maths Chapter 11

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 11.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 11.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 11.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 11.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 11.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 11.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the first question of exercise 11.2 Class 12 Maths about?

The question is to show three lines with given direction cosines are perpendicular

2. Write down the condition for lines with direction cosines l1, m1, n1 and l2, m2 and n2 to be perpendicular?

l1l2+m1m2+n1n2=0

3. How many questions are solved in the NCERT solutions for Class 12 Maths chapter 11 exercise 11.2?

17 questions are solved in the exercise 11.2 Class 12 Maths

4. Which questions depict the concept of the angle between lines?

Question 10 to 13 of Class 12th Maths chapter 11 exercise 11.2

5. Give the topic covered in questions 14 to 17?

Questions 14 to 17 covers the concepts of the shortest distance between two lines.

6. What is the topic discussed after exercise 11.2?

The topic plane is discussed after exercise 11.2

7. What are the main topics covered in Class 12 Maths exercise 11.2?

The main topics covered are equations of lines in three dimensions, angles between lines and least distance between lines.

8. How many exercises are discussed in chapter three dimensional geometry?

4 exercises including the miscellaneous. 

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Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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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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