Careers360 Logo
School
Search Colleges, Exams, Schools & more
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.

PTE Registrations 2024

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

Apply

Access NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Download PDF

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.

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 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.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

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 Question (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. 

Also Read

Articles

Upcoming School Exams

National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

International General Knowledge Olympiad

Exam Date:19 September,2024 - 19 September,2024

Unified Cyber Olympiad

Exam Date:20 September,2024 - 20 September,2024

National Institute of Open Schooling 10th examination

Exam Date:20 September,2024 - 07 October,2024

View All School Exams

Certifications By Top Providers

Edx
 1113 courses
Coursera
 804 courses
Udemy
 394 courses
Futurelearn
 222 courses
Harvard University, Cambridge
 98 courses
IBM
 85 courses

Explore Top Universities Across Globe

University of Essex, Colchester
  Wivenhoe Park Colchester CO4 3SQ
University College London, London
  Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
  Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
  Beacon House, Queens Road, Bristol, BS8 1QU
University of Delaware, Newark
  Newark, DE 19716
University of Nottingham, Nottingham
  University Park, Nottingham NG7 2RD

Related E-books & Sample Papers

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer
Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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 help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
View All

Popular CBSE Class 12th Questions

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
Read More

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

 
Read More

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
Read More

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
Read More

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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

Colleges After 12th

Applications for Admissions are open.

Tallentex 2025 - ALLEN's Talent Encouragement Exam
Tallentex 2025 - ALLEN's Talent Encouragement Exam
Apply

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

PTE Registrations 2024
PTE Registrations 2024
Apply

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

JEE Main high scoring chapters and topics
JEE Main high scoring chapters and topics
Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

View All Application Forms
News and Notifications
CBSE sample paper 2024-25, marking scheme out for skill education subjects
September 02, 2024 - 12:13 PM IST
CBSE to overhaul internal assessment system, develop framework with British Council
August 28, 2024 - 04:58 PM IST
65 lakh students didn't clear Class 10, 12 board exams last year: Education ministry
August 21, 2024 - 10:24 PM IST
CBSE Compartment Result 2024: Marks verification, answer book photocopy, revaluation dates out
August 06, 2024 - 12:37 PM IST
CBSE Class 12 compartment results 2024 declared at cbseresults.nic.in; 29.78% pass
August 02, 2024 - 04:46 PM IST
CBSE best, Kerala board worst in curriculum, infrastructure: NCERT PARAKH report
July 30, 2024 - 04:48 PM IST
Assign 40% weightage to Class 12 exam in final result; have ‘on-demand exams’: NCERT PARAKH report
July 29, 2024 - 07:54 PM IST

Stay up-to date with CBSE Class 12th News

Back to top