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

CBSE Class 12 Maths Chapter 11 Other Study Materials

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.