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

Last updated date: 13th Jul 2024
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## NCERT Solutions for Maths Class 12 Chapter 11 Exercise 11.2 - FREE PDF Download

Exercise 11.2 Class 12 of NCERT Solutions Maths Chapter 11 - Three Dimensional Geometry covers several key topics. These include the equation of a line in space, forming an equation of a line through a given point and parallel to a vector, and the equation of a line passing through two points. Additionally, it addresses the angle between two lines, the shortest distance between two lines, the distance between skew lines, and the distance between parallel lines.

Table of Content
1. NCERT Solutions for Maths Class 12 Chapter 11 Exercise 11.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 12 | Vedantu
3. Formulas Used in Class 12 Chapter 11 Exercise 11.2
4. NCERT Solutions for Class 12 Maths Chapter 11- Three Dimensional Geometry (Exercise 11.2)
5. Class 12 Maths Chapter 11: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 11 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Solving problems in this exercise is crucial for gaining a deep understanding of these concepts. Vedantu's NCERT Solutions offer detailed explanations and step-by-step methods to help students master in ex 11.2 class 12 and excel in their exams.

## Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 12 | Vedantu

• This exercise explains the fundamental concepts of three-dimensional geometry, focusing on equations of lines, angles, and distances in space.

• The equation of a line in space is represented using vector forms, providing a clear understanding of lines in three dimensions.

• The equation of a line through a given point and parallel to a given vector is derived, showing the relationship between points and directions.

• The cartesian form of a line's equation is derived from its vector form, making it easier to solve geometrical problems in coordinate systems.

• The angle between two lines is calculated using their directional cosines, helping to determine their spatial relationship.

• The shortest distance between two lines is determined, providing insight into the minimal separation in space.

• In Three Dimensional Geometry there are 7 fully solved questions of Class 12th Maths, Chapter 11 Exercise 11.2.

## Formulas Used in Class 12 Chapter 11 Exercise 11.2

• Equation of a Line in Vector Form: $r=a+\lambda b$

• Equation of a Line in Cartesian Form: $\frac{x-x_{1}}{a}= \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}$

• Angle Between Two Lines: $cos\theta =\frac{b_{1}\cdot b_{2}}{\left | b_{1} \right |\left | b_{2} \right |}$

• Shortest Distance Between Two Skew Lines: $Distance = \frac{\left | a_{2}-a_{1}\cdot \left ( b_{1}\times b_{2} \right ) \right |}{\left | b_{1}\times b_{2} \right |}$

• Distance Between Two Parallel Lines: $Distance=\frac{\left | d_{1}-d_{2} \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}$

Competitive Exams after 12th Science

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

### Exercise 11.2

1. Show that the three lines with direction cosines  $\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13};\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13};\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ are mutually perpendicular.

Ans: Two lines with direction cosines $\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)$ and $\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)$ are perpendicular to each other if

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$

(i) For the lines with direction cosines

$\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}$ and $\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13}$, we obtain

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{12}{13}\times \dfrac{4}{13}+\dfrac{-3}{13}\times \dfrac{12}{13}+\dfrac{-4}{13}\times \dfrac{3}{13} \\$

$=\dfrac{48}{169}-\dfrac{36}{169}-\dfrac{12}{169} \\$

$=0 \\$

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines

$\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13}$ and $\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$, we obtain

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{4}{13}\times \dfrac{-3}{13}+\dfrac{12}{13}\times \dfrac{-4}{13}+\dfrac{3}{13}\times \dfrac{12}{13} \\$

$=\dfrac{12}{169}-\dfrac{48}{169}+\dfrac{36}{169} \\$

$=0 \\$

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines

$\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ and $\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}$, we obtain

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{3}{13}\times \dfrac{12}{13}+\dfrac{-4}{13}\times \dfrac{-3}{13}+\dfrac{12}{13}\times \dfrac{-4}{13} \\$

$=\dfrac{36}{169}+\dfrac{12}{169}-\dfrac{48}{169} \\$

$=0 \\$

Therefore, the lines are perpendicular.

2. Show that the line through the points $\left( \mathbf{1},-\mathbf{1},\mathbf{2} \right)$ and $\left( \mathbf{3},\mathbf{4},-\mathbf{2} \right)$ is perpendicular to the line through the points $\left( \mathbf{0},\mathbf{3},\mathbf{2} \right)$ and $\left( \mathbf{3},\mathbf{5},\mathbf{6} \right)$.

Ans: Let AB be the line joining the points, $\left( 1,-1,2 \right)$ and $\left( 3,4,-2 \right)$,

and CD be the line through the points $\left( 0,3,2 \right)$ and $\left( 3,5,6 \right)$.

The direction ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ of AB are

$\left( 3-1 \right),\left( 4\left( -1 \right) \right),\left( -2-2 \right)$ i.e.,

$2,5,-4$

The direction ratios ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ of CD are

$\left( 3-0 \right),\left( 5-3 \right),\left( 6-2 \right)$ i.e.,

$3,2,4$

AB and CD will be perpendicular to each other if

${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$

$\left( 2\times 3 \right)+\left( 5\times 2 \right)+\left( -4\times 4 \right) \\$

$=6+10-16 \\$

$=0 \\$

Therefore, AB and CD are perpendicular to each other.

3. Show that the line through the points $\left( \mathbf{4},\mathbf{7},\mathbf{8} \right)$, $\left( \mathbf{2},\mathbf{3},\mathbf{4} \right)$ is parallel to the line through the points $\left( -\mathbf{1},-\mathbf{2},\mathbf{1} \right)$, $\left( \mathbf{1},\mathbf{2},\mathbf{5} \right)$.

Ans: Let AB be the line through the points $\left( 4,7,8 \right)$and $\left( 2,3,4 \right)$,

CD be the line through the points, $\left( -1,-2,1 \right)$and $\left( 1,2,5 \right)$.

The directions ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$, of AB are

$\left( 2-4 \right),\left( 3-7 \right),\left( 4-8 \right)$ i.e.,

$-2,-4,-4$

The direction ratios ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ of CD are

$\left( 1-\left( -1 \right) \right),\left( 2-\left( -2 \right) \right),\left( 5-1 \right)$ i.e.,

$2,4,4$

AB will be parallel to CD, if

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{-2}{2}=-1$…(1)

$\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-4}{4}=-1$…(2)

$\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{4}=-1$…(3)

From Equations (1), (2) and (3), we get

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=-1$

Thus, AB is parallel to CD.

4. Find the equation of the line which passes through point $\left( \mathbf{1},\mathbf{2},\mathbf{3} \right)$ and is parallel to the vector $3\hat{i}+2\hat{j}-2\hat{k}$.

Ans: It is given that the line passes through the point A $\left( 1,2,3 \right)$. Therefore, the position vector through A is $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$

Also, $\vec{b}=3\hat{i}+2\hat{j}-2\hat{k}$

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$ where is $\vec{r}$ a constant.

$\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( 3\hat{i}+2\hat{j}-2\hat{k} \right)$

This is the required equation of the line.

5. Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

Ans: It is given that the line passes through the point with positive vector

$\vec{a}=2\hat{i}-\hat{j}+4\hat{k}$

$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$

$\vec{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda \left( \hat{i}+2\hat{j}-\hat{k} \right)$

This is the required equation in the vector form

$\vec{r}=\left( 2+\lambda \right)\hat{i}+\left( 2\lambda -1 \right)\hat{j}+\left( 4-\lambda \right)\hat{k}$

Eliminating $\lambda$, we obtain the Cartesian form of equation as

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

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

6. Find the Cartesian equation of the line which passes through the point $\left( -\mathbf{2},\mathbf{4},-\mathbf{5} \right)$ and parallel to the line given by $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$.

Ans: It is given that the line passes through the point $(-2,4,-5)$ and is parallel to $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$

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

The required line is parallel to $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$

Therefore, its direction ratios are $3k,5k,6k$, when $k\ne 0$

It is known that the equation of the line through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and with direction ratios $\left( a,b,c \right)$, is given by

$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$

Therefore, the equation of the required line is

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

$\Rightarrow \dfrac{x+2}{3}=\dfrac{y-4}{5}=\dfrac{z+5}{6}=k \\$

7. The Cartesian equation of a line is $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$. Write its vector form.

Ans: The Cartesian equation of the line is $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$.

The given line passes through the point $\left( 5,-4,6 \right)$. The positive vector of this point is $\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}$

Also, the direction ratios of the given line are $3,7,2$.

This means that the line is in the direction of the vector $\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}$

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$

$\vec{r}=\left( 5\hat{i}-4\hat{j}+6\hat{k} \right)+\lambda \left( 3\hat{i}+7\hat{j}+2\hat{k} \right)$

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

8. Find the angle between the following pairs of lines:

(i) $\vec{r}=\left( 2\hat{i}-5\hat{j}+1\hat{k} \right)+\lambda \left( 3\hat{i}-2\hat{j}+6\hat{k} \right)$ and $\vec{r}=\left( 7\hat{i}-6\hat{k} \right)+\mu \left( \hat{i}+2\hat{j}+2\hat{k} \right)$

(ii) $\vec{r}=\left( 3\hat{i}+\hat{j}-2\hat{k} \right)+\lambda \left( \hat{i}-\hat{j}-2\hat{k} \right)$ and $\vec{r}=\left( 2\hat{i}-\hat{j}-56\hat{k} \right)+\mu \left( 3\hat{i}-5\hat{j}-4\hat{k} \right)$

Ans: (i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=3\hat{i}-2\hat{j}+6\hat{k}$ and ${{\vec{b}}_{2}}=\hat{i}+2\hat{j}+2\hat{k}$, respectively.

$|{{b}_{1}}|=\sqrt{{{3}^{2}}+{{2}^{2}}+{{6}^{2}}}=7 \\$

$|{{b}_{2}}|=\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}=3 \\$

${{b}_{1}}\cdot {{b}_{2}}=\left( 3\hat{i}-2\hat{j}+6\hat{k} \right)\cdot \left( \hat{i}+2\hat{j}+2\hat{k} \right) \\$

$=3\times 1+2\times 2+6\times 2 \\$

$=3+4+12 \\$

$=19 \\$

$\cos Q=\dfrac{19}{7\times 3} \\$

$\Rightarrow Q={{\cos }^{-1}}\left( \dfrac{19}{21} \right) \\$

(ii) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=\hat{i}-\hat{j}-2\hat{k}$ and ${{\vec{b}}_{2}}=3\hat{i}-5\hat{j}-4\hat{k}$, respectively.

$|{{b}_{1}}|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{6} \\$

$|{{b}_{2}}|=\sqrt{{{3}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{50}=5\sqrt{2} \\$

${{b}_{1}}\cdot {{b}_{2}}=\left( \hat{i}-\hat{j}-2\hat{k} \right)\cdot \left( 3\hat{i}-5\hat{j}-4\hat{k} \right) \\$

$=1\times 3+\left( -1 \right)\times \left( -5 \right)+\left( -2 \right)\times \left( -4 \right) \\$

$=3+5+8 \\$

$=16 \\$

$\cos Q=\dfrac{16}{\sqrt{6}\times 5\sqrt{2}} \\$

$\Rightarrow Q={{\cos }^{-1}}\left( \dfrac{8}{5\sqrt{3}} \right) \\$

9. Find the angle between following pairs of lines.

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

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

Ans: (i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=2\hat{i}+5\hat{j}-3\hat{k}$ and ${{\vec{b}}_{2}}=-\hat{i}+8\hat{j}+4\hat{k}$, respectively.

$|{{b}_{1}}|=\sqrt{{{2}^{2}}+{{5}^{2}}+{{\left( -3 \right)}^{2}}}=\sqrt{38} \\$

$|{{b}_{2}}|=\sqrt{{{\left( -1 \right)}^{2}}+{{8}^{2}}+{{4}^{2}}}=9 \\$

${{b}_{1}}\cdot {{b}_{2}}=\left( 2\hat{i}+5\hat{j}-3\hat{k} \right)\cdot \left( -\hat{i}+8\hat{j}+4\hat{k} \right) \\$

$=2\times \left( -1 \right)+5\times 8+\left( -3 \right)\times 4 \\$

$=-2+40-12 \\$

$=26 \\$

$\cos Q=\dfrac{26}{9\sqrt{38}} \\$

$\Rightarrow Q={{\cos }^{-1}}\left( \dfrac{26}{9\sqrt{38}} \right) \\$

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=2\hat{i}+2\hat{j}+\hat{k}$ and ${{\vec{b}}_{2}}=8\hat{i}+\hat{j}+8\hat{k}$, respectively.

$|{{b}_{1}}|=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}=3 \\$

$|{{b}_{2}}|=\sqrt{{{4}^{2}}+{{1}^{2}}+{{8}^{2}}}=9 \\$

${{b}_{1}}\cdot {{b}_{2}}=\left( 2\hat{i}+2\hat{j}+\hat{k} \right)\cdot \left( 4\hat{i}+\hat{j}+8\hat{k} \right) \\$

$=2\times 4+2\times 1+1\times 8 \\$

$=8+2+8 \\$

$=18 \\$

$\cos Q=\dfrac{18}{3\times 9} \\$

$\Rightarrow Q={{\cos }^{-1}}\left( \dfrac{2}{3} \right) \\$

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

Ans: The given equations can be written in the standard form as

$\dfrac{x-1}{-3}=\dfrac{y-2}{\dfrac{2p}{7}}=\dfrac{z-3}{2}$ and

$\dfrac{x-1}{-\dfrac{3p}{7}}=\dfrac{y-5}{1}=\dfrac{z-6}{-5}$

The direction ratios of the lines are

$-3,\dfrac{2p}{7},2$ and

$-\dfrac{3p}{7},1,-5$

Two lines will be perpendicular to each other if

${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$

$\left( -3\times \dfrac{-3p}{7} \right)+\left( \dfrac{2p}{7}\times 1 \right)+\left( 2\times \left( -5 \right) \right)=0 \\$

$\Rightarrow \dfrac{9p}{7}+\dfrac{2p}{7}-10=0 \\$

$\Rightarrow 9p+2p-70=0 \\$

$\Rightarrow 11p-70=0 \\$

$\Rightarrow 11p=70 \\$

$\Rightarrow p=\dfrac{70}{11} \\$

Thus, the value of  $p=\dfrac{70}{11}$.

11. Show that the lines $\dfrac{x-5}{7}=\dfrac{y+2}{-5}=\dfrac{z}{1}$ and $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ are perpendicular to each other.

Ans: We are given the equations of lines as

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

The direction ratios of the given lines are

$7,-5,1$ and $1,2,3$

Two lines will be perpendicular to each other if

${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$

$7\times 1+\left( -5 \right)\times 2+1\times 3 \\$

$=7-10+3 \\$

$=0 \\$

Therefore, the given lines are perpendicular to each other.

12. Find the shortest distance between the lines $\vec{r}=\left( \hat{i}+2\hat{j}+\hat{k} \right)+\lambda \left( \hat{i}-\hat{j}+\hat{k} \right)$ and $\vec{r}=\left( 2\hat{i}-\hat{j}-\hat{k} \right)+\mu \left( 2\hat{i}+\hat{j}+2\hat{k} \right)$ .

Ans: The equations of given lines are

$\vec{r}=\left( \hat{i}+2\hat{j}+\hat{k} \right)+\lambda \left( \hat{i}-\hat{j}+\hat{k} \right)$ and $\vec{r}=\left( 2\hat{i}-\hat{j}-\hat{k} \right)+\mu \left( 2\hat{i}+\hat{j}+2\hat{k} \right)$.

It is known that the shortest distance between the lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{1}}+\mu {{\vec{b}}_{2}}$ is given by

$d=\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)\cdot \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$…(1)

On comparing the equations, we obtain,

${{{\vec{a}}}_{1}}=\left( \hat{i}+2\hat{j}+\hat{k} \right) \\$

${{{\vec{b}}}_{1}}=\left( \hat{i}-\hat{j}+\hat{k} \right) \\$

${{{\vec{a}}}_{2}}=\left( 2\hat{i}-\hat{j}-\hat{k} \right) \\$

${{{\vec{b}}}_{2}}=\left( 2\hat{i}+\hat{j}+2\hat{k} \right) \\$

${{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}=\left( 2\hat{i}-\hat{j}-\hat{k} \right)-\left( \hat{i}+2\hat{j}+\hat{k} \right) \\$

$\left( \hat{i}-3\hat{j}-2\hat{k} \right) \\$

\begin{align} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \\ \end{matrix} \right| \\ & =\left( -2-1 \right)\hat{i}-\left( 2-2 \right)\hat{j}+\left( 1+2 \right)\hat{k} \\ & =-3\hat{i}+3\hat{k} \\ \end{align}

$\Rightarrow \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{\left( -3 \right)}^{2}}+{{3}^{2}}} \\$

$=\sqrt{9+9}=\sqrt{18} \\$

$=3\sqrt{2} \\$

Substituting all values obtained in equation (1),

$d=\left| \dfrac{\left( -3\hat{i}+3\hat{k} \right)\cdot \left( \hat{i}-3\hat{j}-2\hat{k} \right)}{3\sqrt{2}} \right| \\$

$=\left| \dfrac{\left( -3 \right)1+3\left( -2 \right)}{3\sqrt{2}} \right| \\$

$=\left| \dfrac{-9}{3\sqrt{2}} \right| \\$

$=\dfrac{3}{\sqrt{2}} \\$

Therefore, the shortest distance between the lines given is $\dfrac{3}{\sqrt{2}}$ units.

13. Find the shortest distance between the lines $\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}$ and $\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}$.

Ans: The given lines are $\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}$ and $\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}$.

It is known that the shortest distance between the lines

$\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ is given by

$d=\left| \dfrac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\sqrt{{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}}} \right|$.…(1)

Comparing the 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, $\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right|$

$=4\left( -16+2 \right)-6\left( 7-1 \right)+8\left( -14+6 \right) \\$

$=-16-36-64 \\$

$=-116 \\$

Also,

$\sqrt{{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}} \\$

$=\sqrt{{{\left( -6+2 \right)}^{2}}+{{\left( 1+7 \right)}^{2}}+{{\left( -14+6 \right)}^{2}}} \\$

$=\sqrt{116} \\$

$=2\sqrt{29} \\$

Substituting all the values in equation (1),

$d=\left| \dfrac{-116}{2\sqrt{29}} \right| \\$

$=\left| \dfrac{-58}{\sqrt{29}} \right| \\$

$=\left| -2\sqrt{29} \right| \\$

$=2\sqrt{29} \\$

Since the distance is always non-negative, the distance between the given lines is $-2\sqrt{29}$ units.

14. Find the shortest distance between the lines whose vector equations are $\vec{r}=\left( \hat{i}+2\hat{j}+3\hat{k} \right)+\lambda \left( \hat{i}-3\hat{j}+2\hat{k} \right)$ and $\vec{r}=\left( 4\hat{i}+5\hat{j}+6\hat{k} \right)+\mu \left( 2\hat{i}+3\hat{j}+\hat{k} \right)$ .

Ans: The equations of given lines are

$\vec{r}=\left( \hat{i}+2\hat{j}+3\hat{k} \right)+\lambda \left( \hat{i}-3\hat{j}+2\hat{k} \right)$ and $\vec{r}=\left( 4\hat{i}+5\hat{j}+6\hat{k} \right)+\mu \left( 2\hat{i}+3\hat{j}+\hat{k} \right)$.

It is known that the shortest distance between the lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{1}}+\mu {{\vec{b}}_{2}}$ is given by

$d=\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)\cdot \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$…(1)

On comparing the equations, we obtain,

${{{\vec{a}}}_{1}}=\left( \hat{i}+2\hat{j}+3\hat{k} \right) \\$

${{{\vec{b}}}_{1}}=\left( \hat{i}-3\hat{j}+2\hat{k} \right) \\$

${{{\vec{a}}}_{2}}=\left( 4\hat{i}+5\hat{j}+6\hat{k} \right) \\$

${{{\vec{b}}}_{2}}=\left( 2\hat{i}+3\hat{j}+\hat{k} \right) \\$

${{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}=\left( 4\hat{i}+5\hat{j}+6\hat{k} \right)-\left( \hat{i}+2\hat{j}+3\hat{k} \right) \\$

$=\left( 3\hat{i}+3\hat{j}+3\hat{k} \right) \\$

\begin{align} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \\ \end{matrix} \right| \\ & =\left( -3-6 \right)\hat{i}-\left( 1-4 \right)\hat{j}+\left( 3+6 \right)\hat{k} \\ & =-9\hat{i}+3\hat{j}+9\hat{k} \\ \end{align}

$\Rightarrow \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{\left( -9 \right)}^{2}}+{{3}^{2}}+{{9}^{2}}} \\$

$=\sqrt{81+9+81}=\sqrt{171} \\$

$=3\sqrt{19} \\$

Substituting all values obtained in equation (1),

$d=\left| \dfrac{\left( -9\hat{i}+3\hat{j}+9\hat{k} \right)\cdot \left( 3\hat{i}+3\hat{j}+3\hat{k} \right)}{3\sqrt{19}} \right| \\$

$=\left| \dfrac{\left( -9 \right)3+3\times 3+9\times 3}{3\sqrt{19}} \right| \\$

$=\left| \dfrac{9}{3\sqrt{19}} \right| \\$

$=\dfrac{3}{\sqrt{19}} \\$

Therefore, the shortest distance between the lines given is $\dfrac{3}{\sqrt{19}}$ units.

15. Find the shortest distance between the lines whose vector equations are

$\vec{r}=(1-t)\hat{i} +(t-2)\hat{j}+(3-2t)\hat{k}$

$\vec{r}=(s+1)\hat{i} +(2s-1)\hat{j}-(2s+1)\hat{k}$

Ans: Firstly, let’s consider the given equations

$\Rightarrow \vec{r}=(1-t)\hat{i} +(t-2)\hat{j}+(3-2t)\hat{k}$

$\vec{r}=\hat{i}-t\hat{i}+t\hat{j}-2\hat{j}+3\hat{k}-2t\hat{k}$

$\vec{r} = \hat{i}-2\hat{j}+3\hat{k}+t(-\hat{i}+\hat{j}-2\hat{k})$

$\Rightarrow \vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$

$\vec{r}=(s+1)\hat{i} +(2s-1)\hat{j}-(2s+1)\hat{k}$

$\vec{r}=s\hat{i}+\hat{i}+2s\hat{j}-\hat{j}-2s\hat{k}-\hat{i}$

$\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2\hat{j}-2\hat{k})$

So, now we need to find the shortest distance between

$\vec{r}=\hat{i}-2\hat{j}+3\hat{k}+t(-\hat{i}+\hat{j}-2\hat{k})$ and $\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2\hat{j}-2\hat{k})$

We know that shortest distance between two lines $\vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}}$ and $\vec{r}=\vec{a_{2}}+ \mu \vec{b_{2}}$

$d=\left | \dfrac{(\vec{b_1}\times \vec{b_2})\cdot (\vec{a_2}\times \vec{a_1}) }{\left | \vec{b_1} \times \vec{b_2} \right |} \right |......(1)$

By comparing the equations we get,

$\vec{a_{1}}=\hat{i}-2\hat{j}+3\hat{k}$ ,  $\vec{b_{1}} =-\hat{i}+\hat{j}-2\hat{k}$

$\vec{a_{2}} =\hat{i}-\hat{j}-\hat{k}$,  $\vec{b_{2}} =\hat{i}+2\hat{j}-2\hat{k}$

Since,

$(X_{1}\hat{i}+Y_{1}\hat{j}+Z_{1}\hat{k})-(X_{2}\hat{i}+Y_{2}\hat{j}+Z_{2}\hat{k})$

$\Rightarrow (X_{1}-X_{2})\hat{i}+(Y_{1}-Y_{2})\hat{j}+(Z_{1}-Z_{2})\hat{k}$

So, $\vec{a_{2}}-\vec{a_{1}}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})=\hat{j}-4\hat{k}$ …..(2)

And,

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

$\begin{vmatrix} \hat{i} & \hat{j}& \hat{k} \\ -1 & 1& -2\\ 1& 2& -2\\ \end{vmatrix}$

$\vec{b_{1}}\times\vec{b_{2}}=2\hat{i}-4\hat{j}-3\hat{k}$ …..(3)

$\Rightarrow \left | \vec{b_{1}}\times\vec{b_{2}} \right |=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=29$

Now multiplying (2) and (3) we get,

$\left ( a_{1}\hat{i} + b_{1}\hat{j} + c_{1}\hat{k} \right )\cdot \left ( a_{2}\hat{i} + b_{2}\hat{j} + c_{2}\hat{k} \right ) = a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$

$(\vec{b_{1}}\times\vec{b_{2}})\cdot (\vec{a_{1}}\times\vec{a_{2}})=(2\hat{i}-4\hat{j}-3\hat{k})\cdot (\hat{j}-4\hat{k})$

=-4+18=8 ….(5)

By substituting all the values in (1), we obtain

The shortest distance between the two lines,

$d=\left | \dfrac{8}{\sqrt{29}} \right |=\dfrac{8}{\sqrt{29}}$

$\therefore$ The shortest distance is $\dfrac{8}{\sqrt{29}}$

## Conclusion

Class 12 Maths Chapter 11 Exercise 11.2 focuses on the equation of a line in space, angles between lines, and the shortest distance between lines. Understanding these concepts is crucial for mastering three-dimensional geometry. Vedantu’s NCERT Solutions offer clear, step-by-step explanations, making complex topics easier to grasp. By practicing these solutions, students can enhance their problem-solving skills and prepare effectively for exams. Rely on Vedantu for comprehensive and accurate solutions to excel in your studies.

## Class 12 Maths Chapter 11: Exercises Breakdown

 Exercise Number of Questions Exercise 11.1 5 Questions with Solutions

## CBSE Class 12 Maths Chapter 11 Other Study Materials

 S.No. Important Links for Chapter 11 Three Dimensional Geometry 1 Class 12 Three Dimensional Geometry Important Questions 2 Class 12 Three Dimensional Geometry Revision Notes 3 Class 12 Three Dimensional Geometry Important Formulas 4 Class 12 Three Dimensional Geometry NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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

1. What are the main concepts on which Class 12 Maths Exercise 11.2 is based?

The Class 12 Maths Exercise 11.2 is based on Three Dimensional Geometry. The main topics which are covered in this chapter are Equation of a Line in Space, Angle between Two Lines and the Shortest distance between Two Lines. The exercise contains problems on these topics and solving these will help students gain a better understanding of all the concepts of this chapter and score well in the exams.

2. Which is the most important question in Exercise 11.2 of 12th Maths?

The most important questions in Exercise 11.2 of Class 12 Maths are Question no. 12 and Question No. 17. These two questions are repeatedly asked in the board exams. Although the entire exercise is important for scoring well on the boards, these two questions must be practised without fail. Students can refer to the Vedantu NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2 and download the free PDF to access the solved problems of this chapter. These PDFs are also available on the Vedantu app at free of cost.

3. Can I score full marks in Class 12 Chapter 11 Maths?

Yes, you can get full marks in Class 12 Maths. All it takes to score full marks in Maths is constant practice. Having all of your concepts clear in this subject is essential since many questions are expected in the board examination from this chapter. This will make you confident and well versed with the chapter and fetch you full marks in the exam. You must also refer to the NCERT Solutions for Chapter 11 of Class 12 Maths.

4. Do I need to practice all the questions provided in Class 12 Maths Exercise 11.2?

Yes, students should solve all the questions in the Class 12 Maths Exercise 11.2.  To get a clear idea about the various problems from this chapter, and answer all the questions accurately, one needs regular practice. The exercise is prepared in such a way that it will help you in both these aspects. For further guidance, you can refer to NCERT Solutions for Chapter 11 Exercise 11.2 of Class 12 Maths on the Vedantu website for free of cost.

5. The NCERT Class 12 Maths Exercise 11.2 Solutions can be viewed only online?

Yes, NCERT Class 12 Maths Exercise 11.2 Solutions can be viewed online as well as downloaded for offline use. You can follow these steps:

• Visit the page NCERT Solutions for Class 12 Maths Chapter 11.

• The page with Vedantu’s NCERT Class 12 Maths Exercise 11.2 Solutions will open up. You will find the option available to download the PDF from this page.

• Click on the Download PDF option to save it on your device. You can use this PDF offline whenever you want.

6. What types of questions are included in class 12 maths chapter 11 exercise 11.2?

Exercise 11.2 features questions on deriving equations of lines in space, calculating angles between lines, and determining the shortest distances between skew lines and between parallel lines.

7. What are the important formulas used in Exercise 11.2 class 12 maths?

Key formulas in ex 11.2 class 12 maths include the vector and cartesian equations of a line, the formula for the angle between two lines, and formulas for calculating the shortest distance between skew lines and the distance between parallel lines.

8. How do you find the shortest distance between two skew lines in ex 11.2 class 12 maths ncert solutions pdf?

The shortest distance between two skew lines is found using the vector cross product and the formula $Distance = \frac{\left | a_{2}-a_{1}\cdot \left ( b_{1}\times b_{2} \right ) \right |}{\left | b_{1}\times b_{2} \right |}$

9. What is the importance of the angle between two lines in class 12 exercise 11.2?

The angle between two lines is crucial for understanding their spatial relationship. It helps in determining whether the lines are parallel, perpendicular, or intersecting at a certain angle. Calculating this angle using the dot product of their direction vectors is essential in solving many geometrical problems and applications in physics and engineering. Knowing the angle also aids in visualizing the lines' orientations in three-dimensional space.

10. How is the distance between parallel lines calculated in ex 11.2 class 12 maths ncert solutions pdf?

The distance between parallel lines is calculated using the formula $Distance=\frac{\left | d_{1}-d_{2} \right |}{\sqrt{a^{2}+b^{2}+c^{2}}}$ where $d_{1}$​ and $d_{2}$​ are the constants in the lines' equations.

11. How does Vedantu’s NCERT Solutions benefit students in mastering class 12 ex 11.2?

Vedantu’s NCERT Solutions of class 12 maths exercise 11.2 offer clear, concise explanations from expert teachers, making it easier for students to grasp difficult concepts and excel in their exams.

12. What are the important formulas used in Class 12 Maths exercise 11.2?

Key formulas include the ex 11.2 class 12 vector and cartesian equations of a line, the formula for the angle between two lines, and formulas for calculating the shortest distance between skew lines and the distance between parallel lines.