NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 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

CBSE Class 12 Maths Chapter 11 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths