Exercise 11.2 (Revised) - Chapter 11 - Three Dimensional Geometry - Ncert Solutions class 12 - Maths
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Ncert Solutions Class 12 Maths: Chapter 11 - Three Dimensional Geometry
Ex 11.2 Question 1.
Show that the three lines with direction cosines
$\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.
Answer.
Given: Direction cosines of three lines are
$
\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}=l_1, m_1, n_1, \frac{4}{13}, \frac{12}{13}, \frac{3}{13}=l_2, m_2, n_2, \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}=l_3, m_3, n_3,
$
For first two lines, $l_1 l_2+m_1 m_2+n_1 n_2=\left(\frac{12}{13}\right)\left(\frac{4}{13}\right)+\left(\frac{-3}{13}\right)\left(\frac{12}{13}\right)+\left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$
$
=\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=\frac{48-36-12}{169}=\frac{0}{169}=0
$
Since, it is 0 , therefore, the first two lines are perpendicular to each other.
For second and third lines, $l_2 l_3+m_2 m_3+n_2 n_3=\left(\frac{4}{13}\right)\left(\frac{3}{13}\right)+\left(\frac{12}{13}\right)\left(\frac{-4}{13}\right)+\left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$
$
=\frac{12}{169}-\frac{48}{169}+\frac{36}{169}=\frac{12-48+36}{169}=\frac{0}{169}=0
$
Since, it is 0 , therefore, second and third lines are also perpendicular to each other.
For First and third lines, $l_1 l_3+m_1 m_3+n_1 n_3=\left(\frac{12}{13}\right)\left(\frac{3}{13}\right)+\left(\frac{-3}{13}\right)\left(\frac{-4}{13}\right)+\left(\frac{-4}{13}\right)\left(\frac{12}{13}\right)$
$
=\frac{36}{169}+\frac{12}{169}-\frac{48}{169}=\frac{36+12-48}{169}=\frac{0}{169}=0
$
Since it is 0 , therefore, first and third lines are also perpendicular to each other.
Hence, given three lines are mutually perpendicular to each other.
Ex 11.2 Question 2.
Show that the line through the points $(1,-1,2),(3,4,-2)$ is perpendicular to the line through the points $(0,3,2)$ and $(3,5,6)$.
Answer.
We know that direction ratios of the line joining the points $\mathrm{A}(1,-1,2)$ and $\mathrm{B}(3,4,-2)$
$
\begin{aligned}
& \text { are } x_2-x_1, y_2-y_1, z_2-z_1 \\
& \Rightarrow 3-1,4-(-1),-2-2 \\
& \Rightarrow 2,5,-4=a_1, b_1, c_1
\end{aligned}
$
Again, direction ratios of the line joining the points $C(0,3,2)$ and $D(3,5,6)$ are
$
\begin{aligned}
& x_2-x_1, y_2-y_1, z_2-z_1 \\
& \Rightarrow 3-0,5-3,6-2 \\
& \Rightarrow 3,2,4=a_2, b_2, c_2 \text { (say) }
\end{aligned}
$
For lines $\mathrm{AB}$ and $\mathrm{CD}, a_1 a_2+b_1 b_2+c_1 c_2=2 \times 3+5 \times 2+(-4) \times 4=6+10-16=0$
Since, it is 0 , therefore, line $A B$ is perpendicular to line $C D$.
Ex 11.2 Question 3.
Show that the line through points $(4,7,8),(2,3,4)$ is parallel to the line through the points $(-1,-2,1),(1,2,5)$.
Answer.
We know that direction ratios of the line joining the points $\mathrm{A}(4,7,8)$ and $\mathrm{B}(2,3,4)$ are
$
\begin{aligned}
& x_2-x_1, y_2-y_1, z_2-z_1 \\
& \Rightarrow 2-4,3-7,4-8 \\
& \Rightarrow-2,-4,-4=a_1, b_1, c_1 \text { (say) }
\end{aligned}
$
Again direction ratios of the line joining the points $C(-1,-2,1)$ and $D(1,2,5)$ are
$
x_2-x_1, y_2-y_1, z_2-z_1
$
$
\begin{aligned}
& \Rightarrow 1-(-1), 2-(-2), 5-1 \\
& \Rightarrow 2,4,4=a_2, b_2, c_2 \text { (say) }
\end{aligned}
$
For the lines $\mathrm{AB}$ and $\mathrm{CD}, \frac{a_1}{a_2}=\frac{-2}{2}, \frac{b_1}{b_2}=\frac{-4}{4}, \frac{c_1}{c_2}=\frac{-4}{4}=-1$
Since, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Therefore, line $\mathrm{AB}$ is parallel to line $\mathrm{CD}$.
Ex 11.2 Question 4.
Find the equation of the line which passes through the point $(1,2,3)$ and is parallel to the vector $3 \hat{i}+2 \hat{j}-2 \hat{k}$
Answer.
A point on the required line is A $(1,2,3)=x_1, y_1, z_1$
$\Rightarrow$ Position vector of a point on the required line is $\vec{a}=\overrightarrow{\mathrm{OA}}=(1,2,3)=\hat{i}+2 \hat{j}+3 \hat{k}$
The required line is parallel to the vector $\vec{b}=3 \hat{i}+2 \hat{j}-2 \hat{k}$
$\therefore$ direction ratios of the required line are coefficient of $\hat{i}, \hat{j}, \hat{k}$ in $\vec{b}$ are
$3,2,-2=a, b, c$
$\therefore$ Vector equation of the required line is
$
\vec{r}=\vec{a}+\lambda \vec{b} \Rightarrow \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})
$
Where $\lambda$ is a real number.
Cartesian equation of this equation is $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}$
Ex 11.2 Question 5.
Find the equation of the line in vector and in Cartesian form that passes through the
point with position vector $2 \hat{i}-\hat{j}+4 \hat{k}$ and is in the direction $\hat{i}+2 \hat{j}-\hat{k}$.
Answer.
Position vector of a point on the required line is $\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}=$
$
(2,-1,4)=\left(x_1, y_1, z_1\right)
$
The required line is in the direction of the vector is $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$
$\Rightarrow$ Direction ratios of required line are coefficients of $\hat{i}, \hat{j}, \hat{k}$ in $\vec{b}=1,2,-1=a, b, c$
$\therefore$ Equation of the required line in vector form is $\vec{r}=\vec{a}+\lambda \vec{b}$
$
\Rightarrow \vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})
$
Where $\lambda$ is a real number.
Cartesian equation of this equation is $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$
Ex 11.2 Question 6.
Find the Cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$.
Answer.
Given: A point on the line is $(-2,4,-5)=\left(x_1, y_1, z_1\right)$
Equation of the given line in Cartesian form is $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
$\therefore$ Direction ratios of the given line are its denominators $3,5,6=a, b, c$
$\therefore$ Equation of the required line is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$
\Rightarrow \frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6}=\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}
$
Ex 11.2 Question 7.
The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$. Write its vector form.
Answer.
Given: The Cartesian equation of the line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}=\lambda$ (say)
$
\begin{aligned}
& \Rightarrow x-5=3 \lambda, y+4=7 \lambda, z-6=2 \lambda \\
& \Rightarrow x=5+3 \lambda, y=-4+7 \lambda, z=6+2 \lambda
\end{aligned}
$
General equation for the required line is $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$
Putting the values of $x, y, z$ in this equation,
$
\begin{aligned}
& \vec{r}=(5+3 \lambda) \hat{i}+(-4+7 \lambda) \hat{j}+(6+2 \lambda) \hat{k}=5 \hat{i}+3 \hat{i}-4 \hat{j}+7 \lambda \hat{j}+6 \hat{k}+2 \lambda \hat{k} \\
& \Rightarrow \vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})[\operatorname{Since} \vec{r}=\vec{a}+\lambda \vec{b}]
\end{aligned}
$
Ex 11.2 Question 10.
Find the angle between the following pairs of lines:
(i) $\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$ and $\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$
(ii) $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-5 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k})$
Answer.
(i) Equation of the first line is $\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})$
Comparing with $(\vec{r}=\vec{a}+\lambda \vec{b})$,
$\overrightarrow{a_1}=2 \hat{i}-5 \hat{j}+\hat{k}$ and $\overrightarrow{b_1}=3 \hat{i}+2 \hat{j}+6 \hat{k}$
(vector $\vec{a}$ is the position vector of a point on line and $\vec{b}$ is a vector along the line)
Again, equation of the second line is $\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$
Comparing with $(\vec{r}=\vec{a}+\mu \vec{b})$,
$\overrightarrow{a_2}=7 \hat{i}-6 \hat{k}$ and $\overrightarrow{b_2}=\hat{i}+2 \hat{j}+2 \hat{k}$
(vector $\vec{a}$ is the position vector of a point on line and $\vec{b}$ is a vector along the line)
Let $\theta$ be the angle between these two lines, then
$
\begin{aligned}
& \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right| \cdot\left|\overrightarrow{b_2}\right|}=\frac{3(1)+2(2)+6(2)}{\sqrt{9+4+36} \sqrt{1+4+4}}=\frac{3+4+12}{\sqrt{49} \sqrt{9}}=\frac{19}{7 \times 3} \\
& \cos \theta=\frac{19}{21} \Rightarrow \theta=\cos ^{-1} \frac{19}{21}
\end{aligned}
$
(ii)Comparing the first and second equations with $(\vec{r}=\vec{a}+\lambda \vec{b})$ and $(\vec{r}=\vec{a}+\mu \vec{b})$ resp.
$
\overrightarrow{b_1}=\hat{i}-\hat{j}-2 \hat{k} \text { and } \overrightarrow{b_2}=3 \hat{i}-5 \hat{j}-4 \hat{k}
$
Let $\theta$ be the angle between these two lines, then
$
\begin{aligned}
& \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right| \cdot\left|\overrightarrow{b_2}\right|}=\frac{1(3)+(-1)(-5)+(-2)(-4)}{\sqrt{1+1+4} \sqrt{9+25+16}}=\frac{3+5+8}{\sqrt{6} \sqrt{50}}=\frac{16}{\sqrt{300}} \\
& \cos \theta=\frac{16}{10 \sqrt{3}}=\frac{8}{5 \sqrt{3}} \Rightarrow \theta=\cos ^{-1} \frac{8}{5 \sqrt{3}} \Rightarrow \theta=\cos ^{-1} \frac{8 \sqrt{3}}{15}
\end{aligned}
$
Ex 11.2 Question 11.
Find the angle between the following pair of lines:
(i) $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
(ii) $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$
Answer.
(i) Given: Equation of first line is $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$
The direction ratios of this line i.e., a vector along the line is
$
\overrightarrow{b_1}=(2,5,-3)=2 \hat{i}+5 \hat{j}-3 \hat{k}
$
Now, equation of second line is $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
The direction ratios of this line i.e., a vector along the line is
$
\overrightarrow{b_2}=(-1,8,4)=-\hat{i}+8 \hat{j}+4 \hat{k}
$
Let $\theta$ be the angle between these two lines, then
$
\begin{aligned}
& \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right| \cdot\left|\overrightarrow{b_2}\right|}=\frac{2(-1)+(5)(8)+(-3)(4)}{\sqrt{4+25+9} \sqrt{1+64+16}}=\frac{-2+40-12}{\sqrt{38} \sqrt{81}}=\frac{26}{9 \sqrt{38}} \\
& \Rightarrow \theta=\cos ^{-1} \frac{26}{9 \sqrt{38}}
\end{aligned}
$
(ii)Given: Equation of first line is $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$
The direction ratios of this line i.e., a vector along the line is
$
\overrightarrow{b_1}=(2,2,1)=2 \hat{i}+2 \hat{j}+\hat{k}
$
Nowequation of second line is $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$
The direction ratios of this line i.e., a vector along the line is
$
\overrightarrow{b_2}=(4,1,8)=4 \hat{i}+\hat{j}+8 \hat{k}
$
Let $\theta$ be the angle between these two lines, then
$
\begin{aligned}
& \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right| \cdot\left|\overrightarrow{b_2}\right|}=\frac{2(4)+(2)(1)+(1)(8)}{\sqrt{4+4+1} \sqrt{16+1+64}}=\frac{8+2+8}{\sqrt{9} \sqrt{81}}=\frac{18}{3 \times 9}=\frac{2}{3} \\
& \Rightarrow \theta=\cos ^{-1} \frac{2}{3}
\end{aligned}
$
Ex 11.2 Question 12.
$\text {Find the values of } p \text { so that the lines } \frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} \text { and }$
$\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.
Answer.
Given: Equation of one line $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} \Rightarrow$
$
\begin{aligned}
& \frac{-(x-1)}{3}=\frac{7(y-2)}{2 p}=\frac{z-3}{2} \\
& \Rightarrow \frac{-(x-1)}{3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2}
\end{aligned}
$
$\therefore$ Direction ratios of this line are $-3, \frac{2 p}{7}, 2=a_1, b_1, c_1$
(say)
Again, equation of another line $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5} \Rightarrow$
$
\begin{aligned}
& \frac{-7(x-1)}{3 p}=\frac{y-5}{1}=\frac{-(z-6)}{5} \\
& \Rightarrow \frac{x-1}{\frac{-3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}
\end{aligned}
$
$\therefore$ Direction ratios of this line are $\frac{-3 p}{7}, 1,-5=a_2, b_2, c_2$
(say)
Since, these two lines are perpendicular.
Therefore, $a_1 a_2+b_1 b_2+c_1 c_2=0$
$\begin{aligned}
& \Rightarrow(-3)\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \Rightarrow \frac{9 p}{7}+\frac{2 p}{7}-10=0 \\
& \Rightarrow \frac{11 p}{7}=10 \Rightarrow p=\frac{70}{11}
\end{aligned}$
Ex 11.2 Question 13.
Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.
Answer.
Equation of one line $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$
$\therefore$ Direction ratios of this line are $7,-5,1=a_1, b_1, c_1$
$
\Rightarrow \overrightarrow{b_1}=7 \hat{i}-5 \hat{j}+\hat{k}
$
Again equation of another line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
$\therefore$ Direction ratios of this line are $1,2,3=a_2, b_2, c_2$
$
\begin{aligned}
& \Rightarrow \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \text { Now } \overrightarrow{b_1} \cdot \overrightarrow{b_2}=a_1 a_2+b_1 b_2+c_1 c_2=7 \times 1+(-5) \times 2+1 \times 3=7-10+3=0
\end{aligned}
$
Hence, the given two lines are perpendicular to each other.
Ex 11.2 Question 14.
Find the shortest distance between the lines $\vec{r}=\hat{i}+2 \hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})$
Answer.
Comparing the given equations with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\lambda \overrightarrow{b_2}$, we get
$
\overrightarrow{a_1}=\hat{i}+2 \hat{j}+\hat{k}, \quad \overrightarrow{b_1}=\hat{i}-\hat{j}+\hat{k} \quad \text { and } \overrightarrow{a_2}=2 \hat{i}-\hat{j}-\hat{k}, \quad \overrightarrow{b_2}=2 \hat{i}+\hat{j}+2 \hat{k}
$
Since, the shortest distance between the two skew lines is given by
$
d=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}
$
$
\begin{aligned}
& \text { Here, } \overrightarrow{a_2}-\overrightarrow{a_1}=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k} \\
& \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 1 \\
2 & 1 & 2
\end{array}\right|=(-2-1) \hat{i}-(2-2) \hat{j}+(1+2) \hat{k}=-3 \hat{i}+3 \hat{k} \\
& \left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-3)^2+(0)^2+(3)^2}=\sqrt{18}=3 \sqrt{2} \\
& \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=(\hat{i}-3 \hat{j}-2 \hat{k}) \cdot(-3 \hat{i}+3 \hat{k})=1 \times(-3)+(-3 \times 0)+(-2 \times 3)=-9
\end{aligned}
$
Putting these values in eq. (i),
Shortest distance $(d)=\frac{|-9|}{3 \sqrt{2}}=\frac{9}{3 \sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3 \sqrt{2}}{2}$
Ex 11.2 Question 15.
Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Answer.
Equation of one line is $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$
Comparing this equation with $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$, we have
$
x_1=-1, y_1=-1, \quad z_1=-1, \quad a_1=7, b_1=-6, c_1=1
$
Again equation of another line is $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Comparing this equation with $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$, we have
$
\begin{aligned}
& x_2=3, y_2=5, z_2=7, a_2=1, b_2=-2, c_2=1 \\
& \therefore\left|\begin{array}{ccc}
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{array}\right|=\left|\begin{array}{ccc}
3+1 & 5+1 & 7+1 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right|
\end{aligned}
$
Expanding by first row $=4(-6+2)-6(7-1)+8(-14+6)=-16-36-64=-116$
$
\begin{aligned}
& \text { And } \sqrt{\left(a_1 b_2-a_2 b_1\right)+\left(b_1 c_2-b_2 c_1\right)+\left(c_1 a_2-c_2 a_1\right)} \\
& =\sqrt{(-14+6)^2+(-6+2)^2+(1-7)^2}=\sqrt{64+16+36}=\sqrt{116} \\
& \therefore \text { Length of shortest distance }=\frac{\left|\begin{array}{ccc}
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{array}\right|}{\sqrt{\left(a_1 b_2-a_2 b_1\right)+\left(b_1 c_2-b_2 c_1\right)+\left(c_1 a_2-c_2 a_1\right)}} \\
& =\frac{-116}{\sqrt{116}}=-\sqrt{116}=\sqrt{116} \text { (numerically) } \\
& =\sqrt{4 \times 29}=2 \sqrt{29} \\
&
\end{aligned}
$
Ex 11.2 Question 16.
Find the shortest distance between the lines whose vector equations are
$
\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+2 \hat{k}) \text { and } \vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})
$
Answer.
Equation of the first line is $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$
Comparing this equation with $\vec{r}=\vec{a}_1+\lambda \overrightarrow{b_1}$,
$
\overrightarrow{a_1}=\hat{i}+2 \hat{j}+3 \hat{k} \text { and } \overrightarrow{b_1}=\hat{i}-3 \hat{j}+2 \hat{k}
$
Again equation of second line $\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$
Comparing this equation with $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$,
$\overrightarrow{a_2}=4 \hat{i}+5 \hat{j}+6 \hat{k}$ and $\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}$
Now shortest distance $(d)=\frac{\left|\left(\overline{a_2}-\overline{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overline{b_2}\right)\right|}{\left|\bar{b}_1 \times \bar{b}_2\right|}$
Here $\overrightarrow{a_2}-\overrightarrow{a_1}=(4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+3 \hat{j}+3 \hat{k}$
$\overline{b_1} \times \overline{b_2}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1\end{array}\right|=(-3-6) \hat{i}-(1-4) \hat{j}+(3+6) \hat{k}=-9 \hat{i}+3 \hat{j}+9 \hat{k}$
$\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{171}=3 \sqrt{19}$
$\left(\bar{a}_2-\bar{a}_1\right) \cdot\left|\bar{b}_1 \times \bar{b}_2\right|=3 \times(-9)+(3 \times 3)+(3 \times 9)=-27+9+27=9$
Putting these values in eq. (i),
Shortest distance $(d)=\frac{|9|}{3 \sqrt{19}}=\frac{9}{3 \sqrt{19}}=\frac{3}{\sqrt{19}}$
Ex 11.2 Question 17.
Find the shortest distance between the lines whose vector equations are
$
\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \text { and } \vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k} \text {. }
$
Answer.
Equation of first line is $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$
$
=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})
$
Comparing this equation with $\overrightarrow{a_1}+t \vec{b}_1$,
$
\overrightarrow{a_1}=\hat{i}-2 \hat{j}+3 \hat{k}, \quad \overrightarrow{b_1}=-\hat{i}+\hat{j}-2 \hat{k}
$
Equation of second line is $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$
$
=s \hat{i}+\hat{i}+2 s \hat{j}-\hat{j}-2 s \hat{k}-\hat{k}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})
$
Comparing this equation with $\overrightarrow{a_2}+s \overrightarrow{b_2}$,
$
\begin{aligned}
& \overrightarrow{a_2}=\hat{i}-\hat{j}-\hat{k}, \quad \overrightarrow{b_2}=\hat{i}+2 \hat{j}-2 \hat{k} \\
& \text { Now Shortest distance }(d)=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}
\end{aligned}
$
Here $\overrightarrow{a_2}-\overrightarrow{a_1}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\hat{j}-4 \hat{k}$
$
\begin{aligned}
& \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{l}
\hat{i} \hat{j} \hat{k} \\
-11-2 \\
12-2
\end{array}\right|(-2+4) \hat{i}-(2+2) \hat{j}+(-2-1) \hat{k} \\
& \left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(2)^2+(-4)^2+(-3)^2}=\sqrt{29} \\
& \left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=(\hat{j}-4 \hat{k}) \cdot(2 \hat{i}-4 \hat{j}-3 \hat{k})=0 \times 2+1 \times(-4)+(-4)(-3)=8
\end{aligned}
$
Putting these values in eq. (i),
Shortest distance $(d)=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}}$
