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

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## Download PDF of NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.2) Exercise 11.2

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

Topics Covered in Class 12 Maths Chapter 11 Exercise 11.2

Equation of a Line in Space

Equation of a line through a given point and parallel to a given vector b.

Equation of a line passing through two given points

Angle between Two Lines

Shortest Distance between Two Lines

Distance between two skew lines

Distance between parallel lines

### 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 vector and the Cartesian equations of the lines that passes through the origin and \[\left( \mathbf{5},-\mathbf{2},\mathbf{3} \right)\].

Ans: The required line passes through the origin. Therefore, its positive vector is given by,

$\vec{a}=0$…(1)

The direction ratios of the line through origin and \[\left( 5,-2,3 \right)\] are

\[\left( 5-0 \right)=5,\left( -2-0 \right)=-2,\left( 30 \right)=3\]

The line is parallel to the vector given by the equation,

$\vec{b}=5\hat{i}-2\hat{j}+3\hat{k}$

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( 0 \right)+\lambda \left( 5\hat{i}-2\hat{j}+3\hat{k} \right) \\ $

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

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-0}{5}=\dfrac{y-0}{-2}=\dfrac{z-0}{3} \\$

$ \Rightarrow \dfrac{x}{5}=\dfrac{y}{-2}=\dfrac{z}{3} \\ $

9. Find the vector and the Cartesian equation of the line that passes through the point \[\left( \mathbf{3},-\mathbf{2},-\mathbf{5} \right),\text{ }\left( \mathbf{3},-\mathbf{2},\mathbf{6} \right)\].

Ans: Let the line passing through the points, P \[\left( 3,-2,-5 \right)\]and Q \[\left( 3,-2,6 \right)\], be PQ.

Since PQ passes through P \[\left( 3,-2,-5 \right)\], its positive vector is given by,

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

The direction ratios of PQ are given by,

\[\left( 3-3 \right)=0,\text{ }\left( -2+2 \right)=0,\text{ }\left( 6+5 \right)=11\]

The equation of the vector in the direction of PQ is

$\vec{b}=0\hat{i}+0\hat{j}+11\hat{k}=11\hat{k}$

$ \vec{r}=\left( 3\hat{i}-2\hat{j}-5\hat{k} \right)+\lambda \left( 0\hat{i}+0\hat{j}+11\hat{k} \right) \\ $

$ \Rightarrow \vec{r}=\left( 3\hat{i}-2\hat{j}-5\hat{k} \right)+\lambda \left( 11\hat{k} \right) \\ $

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

\[\dfrac{x-3}{0}=\dfrac{y+2}{0}=\dfrac{z+5}{11}\]

10. 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) \\ $

11. 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 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) \\ $

12. 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 $p=\dfrac{70}{11}$.

13. 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.

14. 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.

15. 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.

16. 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.

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

Ans: The equations of given lines are

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

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}+3\hat{k} \right) \\ $

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

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

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

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

$ =\left( \hat{j}-4\hat{k} \right) \\ $

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

$ \Rightarrow \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{2}^{2}}+{{\left( -4 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\$

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

$ =\sqrt{29} \\$

Substituting all values obtained in equation (1),

$ d=\left| \dfrac{\left( 2\hat{i}-4\hat{j}-3\hat{k} \right)\cdot \left( \hat{j}-4\hat{k} \right)}{\sqrt{29}} \right| \\ $

$ =\left| \dfrac{-4\times 1+\left( -3 \right)\times \left( -4 \right)}{\sqrt{29}} \right| \\ $

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

Therefore, the shortest distance between the lines given is \[\dfrac{8}{\sqrt{29}}\] units.

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

Opting for the NCERT solutions for Ex 11.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 11 Exercise 11.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Important Features of NCERT Solutions for Class 12 Maths Chapter 11 - Three Dimensional Geometry

Three-dimensional geometry is the study of the properties of lines, points, and various solid shapes in three-dimensional coordinate systems.

Some of the major concepts in this exercise are explained using the concepts of cartesian and vector equations of a line in three-dimensional geometry.

### NCERT Solutions for Class 12 Maths PDF Download

Besides these NCERT solutions for Class 12 Maths Chapter 11 Exercise 11.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 11 Exercise 11.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

Chapter 11 - Three-Dimensional Geometry |

## Exercise 11.1 |

## Exercise 11.2 |

## Exercise 11.3 |

## FAQs on NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.2) 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.