Chapter 11: Three-dimensional Geometry

Solution: Recall that the direction cosines of a line making angles $\alpha, \beta, \gamma$ with the $x, y$ and $z$-axes respectively are $\boldsymbol{\operatorname { c o s }} \alpha, \boldsymbol{\operatorname { c o s }} \beta, \boldsymbol{\operatorname { c o s }} \gamma$. Here $\alpha=90^{\circ}, \beta=135^{\circ}$ and $\gamma=45^{\circ}$.
Therefore, direction cosines of the required line are $\cos 90^{\circ}$, $\cos 135^{\circ}$ and $\cos 45^{\circ}=0, \frac{-1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$.
$
\left[\cos 135^{\circ}=\cos \left(180^{\circ}-45^{\circ}\right)=-\cos 45^{\circ}=\frac{-1}{\sqrt{2}}\right]
$

Result. $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$.

Solution: Let a line make equal angles $\alpha, \alpha, \alpha$ with the co-ordinate axes.
∴ Direction cosines of the line are $\cos \alpha, \cos \alpha, \cos \alpha \ldots(i)$
$\therefore \quad \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \quad\left[\because \quad \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right]$
$\Rightarrow 3 \cos ^2 \alpha=1 \Rightarrow \cos ^2 \alpha=\frac{1}{3} \Rightarrow \cos \alpha= \pm \sqrt{\frac{1}{3}}= \pm \frac{1}{\sqrt{3}}$
Substituting $\cos \alpha= \pm \frac{1}{\sqrt{3}}$ in (i), direction cosines of the required line making equal angles with the co-ordinate axes are
$
\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} .
$

Very Important Remark. Therefore, direction cosines of a line making equal angles with the co-ordinate axes in the positive (i.e., first) octant are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

Solution: Recall that if $a, b, c$ are direction ratios of a line, then the direction cosines of the line are
$
\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}
$

Here, direction ratios of the line are
$
\begin{aligned}
-18, & 12,-4=a, b, c \\
\therefore \sqrt{a^2+b^2+c^2} & =\sqrt{(-18)^2+(12)^2+(-4)^2}=\sqrt{324+144+16} \\
& =\sqrt{484}=22
\end{aligned}
$

Substituting these values in (i), direction cosines of the required line are
$
\frac{-18}{22}, \frac{12}{22}, \frac{-4}{22}=\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11} .
$

Solution: The given points are $\mathrm{A}(2,3,4), \mathrm{B}(-1,-2,1)$ and $\mathrm{C}(5,8,7)$.
∴ Direction ratios of the line joining A and B are
$
\begin{array}{llr}
& -1-2,-2-3,1-4 & \mid x_2-x_1, y_2-y_1, z_2-z_1 \\
i . e ., & -3,-5,-3 &
\end{array}
$

Next direction ratios of the line joining B and C are
$
\begin{aligned}
5-(-1), 8-(-2), 7-1 & =6,10,6 \\
& =a_2, b_2, c_2(\text { say })
\end{aligned}
$

From (i) and (ii) direction ratios of AB and BC are proportional
$
\text { i.e., } \quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \quad\left[\because \frac{-3}{6}=\frac{-5}{10}, \frac{-3}{6}\left(\text { each }=\frac{-1}{2}\right)\right]
$

Therefore, AB is parallel to BC . But point B is common to both AB and BC . Hence, points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear.

Solution: Direction ratios of line AB are $-1-3,1-5,2-(-4)$
i.e., $\quad-4,-4,6 \quad \mid x_2-x_1, y_2-y_1, z_2-z_1$
Dividing each by $\sqrt{a^2+b^2+c^2}=\sqrt{(-4)^2+(-4)^2+6^2}$
$
=\sqrt{16+16+36}=\sqrt{68}=\sqrt{4 \times 17}=2 \sqrt{17} .
$
direction cosines of line AB are
$
\begin{aligned}
& \frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}} \\
& \text { i.e., } \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}
\end{aligned}
$

$(-5,-5,-2)$

Direction ratios of line BC are
$
-5-(-1),-5-1,-2-2=-4,-6,-4
$

Dividing each by $\sqrt{(-4)^2+(-6)^2+(-4)^2}=\sqrt{16+36+16}$
$
=\sqrt{68}=\sqrt{4 \times 17}=2 \sqrt{17}
$

Direction cosines of line BC are $\frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}$
i.e., $\quad \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$
Direction ratios of line CA are
$
3-(-5), 5-(-5),-4-(-2)=8,10,-2
$

Dividing each by $\sqrt{(8)^2+(10)^2+(-2)^2}=\sqrt{64+100+4}$
$
=\sqrt{168}=\sqrt{4 \times 42}=2 \sqrt{42} .
$

Direction ratios of line CA are
$
\frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{-2}{2 \sqrt{42}}=\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}
$

Note. If $l, m, n$ are direction cosines of a line, then $-l,-m,-n$ are also direction cosines of the same line.

Solution: Given: Direction cosines of three lines are
$
\begin{aligned}
& \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ;=l_1, m_1, n_1, \quad \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ;=l_2, m_2, n_2 \\
& \text { and } \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}=l_3, m_3, n_3
\end{aligned}
$





For first two lines;
$
\begin{aligned}
& l_1 l_2+m_1 m_2+n_1 n_2=\frac{12}{13}\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
\end{aligned}
$
∴ The first two lines are perpendicular to each other.
For second and third line,
$
\begin{aligned}
& l_2 l_3+m_2 m_3+n_2 n_3 \\
= & \frac{4}{13}\left(\frac{3}{13}\right)+\frac{12}{13}\left(\frac{-4}{13}\right)+\frac{3}{13}\left(\frac{12}{13}\right)=\frac{12}{169}-\frac{48}{169}+\frac{36}{169} \\
= & \frac{12-48+36}{169}=\frac{0}{169}=0
\end{aligned}
$
∴ Second and third lines are perpendicular to each other.
For first and third line,
$
\begin{aligned}
& l_1 l_3+m_1 m_3+n_1 n_3 \\
= & \frac{12}{13}\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
\end{aligned}
$
∴ First and third line are also perpendicular to each other.
∴ The three given lines are mutually perpendicular.

Solution: Recall that the direction ratios of the line joining the points
$\mathrm{A}(1,-1,2)$ and $\mathrm{B}(3,4,-2)$ are $x_2-x_1, y_2-y_1, z_2-z_1$ i.e., $3-1,4-(-1),-2-2=2,5,-4=a_1, b_1, c_1$ (say)
Next, direction ratios of the line joining the points $\mathrm{C}(0,3,2)$ and $\mathrm{D}(3,5,6)$ are $x_2-x_1, y_2-y_1, z_2-z_1$ i.e., $\quad 3-0,5-3,6-2=3,2,4=a_2, b_2, c_2$ (say)
For these lines AB and CD ,
$
\begin{aligned}
a_1 a_2+b_1 b_2+c_1 c_2 & =2(3)+5(2)+(-4)(4) \\
& =6+10-16=0
\end{aligned}
$
∴ Given line AB is perpendicular to given line CD .

Solution: Recall that the direction ratios of the line joining the points $\mathrm{A}(4,7,8)$ and $\mathrm{B}(2,3,4)$ are $x_2-x_1, y_2-y_1, z_2-z_1$ i.e., $\quad 2-4,3-7,4-8 \quad$ i.e., $\quad-2,-4,-4=a_1, b_1, c_1$ (say)
Next, direction ratios of the line joining the points $\mathrm{C}(-1,-2,1)$ and $\mathrm{D}(1,2,5)$ are $1-(-1), 2-(-2), 5-1=2,4,4=a_2, b_2, c_2$ (say)
For these lines AB and CD ,




$
\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \quad\left(\text { as } \frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}(=-1 \text { each })\right)
$
∴ Given line AB is parallel to given line CD .

Solution: A point on the required line is
$
\mathrm{A}(1,2,3)=\left(x_1, y_1, z_1\right)
$
i.e., 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 $\vec{b}=3 \hat{i}+2 \hat{j}-2 \hat{k}$ (and hence direction ratios of the required line are coefficient of
$
\hat{i}, \hat{j}, \hat{k} \text { in } \vec{b} \quad \text { i.e., } \quad 3,2,-2=a, b, c)
$
∴ Vector equation of required line is
$
\overrightarrow{\boldsymbol{r}}=\overrightarrow{\boldsymbol{a}}+\lambda \overrightarrow{\boldsymbol{b}} \quad \text { i.e., } \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.
Remark. Also cartesian equation of the required line in this Q. No. 4 is
$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \text { i.e., } \frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}
$

Solution: 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
$
\vec{b}=\hat{i}+2 \hat{j}-\hat{k}
$
( ⇒ direction ratios of required line are coefficients of $\hat{i}, \hat{j}, \hat{k}$ in $\vec{b}$ i.e., $1,2,-1=a, b, c$ )
∴ Equation of the required line in vector form is $\overrightarrow{\boldsymbol{r}}=\overrightarrow{\boldsymbol{a}}+\lambda \overrightarrow{\boldsymbol{b}}$
$
\text { i.e., } \quad \vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})
$
where $\lambda$ is a real number and equation of line in cartesian form is
$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \quad \text { i.e., } \quad \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1} .
$

Solution: Given: A point on the required line is $(-2,4,-5)=\left(x_1, y_1, z_1\right)$.
Equations of the given line in cartesian form are
$
\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}
$
(It is standard form because coefficients
of $x, y, z$ are unity each)
∴ Direction ratios (D.R.’s) of the given line are its denominators $3,5,6$ and hence d.r.’s of the required parallel line are also 3,5 , $6=a, b, c$.
∴ Equations of the required line are
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \quad$ i.e. $\quad \frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6}$
i.e., $\quad \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$.

Solution: Given: The cartesian equation of a line is
$
\begin{aligned}
& \quad \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \\
& \text { ie } \frac{x-5}{3}=\frac{y-(-4)}{7}=\frac{z-6}{2}
\end{aligned}
$
comparing the given equation with the standard form
$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
$
we have $x_1=5, y_1=-4, z_1=6 ; a=3, b=7, c=2$
Therefore, the given line passes through the point
$
\vec{a}=\left(x_1, y_1, z_1\right)=(5,-4,6)=5 \hat{i}-4 \hat{j}+6 \hat{k}
$
and is parallel (or collinear) with the vector
$
\vec{b}=a \hat{i}+b \hat{j}+c \hat{k}=3 \hat{i}+7 \hat{j}+2 \hat{k}
$
∴ Vector equation of the given line is $\vec{r}=\vec{a}+\lambda \vec{b}$
i.e $\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$

Solution: (i) Given: Equation of one line is
$
\vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})
$

Comparing with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$,
$\overrightarrow{a_1}=2 \hat{i}-5 \hat{j}+\hat{k}$ and a vector along the line is
$
\overrightarrow{b_1}=3 \hat{i}+2 \hat{j}+6 \hat{k}
$
(Note that the vector $\overrightarrow{a_1}$ is the position vector of a point on the line and not a vector along the line).
Given: Equation of second line is
$
\vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k})
$

Comparing with $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ we have
$\overrightarrow{a_2}=7 \hat{i}-6 \hat{k}$ and a vector along the second line is
$
\overrightarrow{b_2}=\hat{i}+2 \hat{j}+2 \hat{k}
$






Let $\theta$ be the angle between the two lines.
$
\begin{aligned}
& \text { Recall that } \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\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}} \\
& \cos \theta=\frac{19}{7(3)}=\frac{19}{21} \quad \therefore \quad \theta=\cos ^{-1} \frac{19}{21}
\end{aligned}
$
(ii) Comparing the equations of the two given lines with
$\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$ we have
$\overrightarrow{b_1}=\hat{i}-\hat{j}-2 \hat{k}$ and $\overrightarrow{b_2}=3 \hat{i}-5 \hat{j}-4 \hat{k}$.
Let $\theta$ be the angle between the two lines
$
\begin{aligned}
& \therefore \quad \cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\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}}=\frac{16}{\sqrt{3 \times 100}}=\frac{16}{10 \sqrt{3}} \\
& \text { or } \cos \theta=\frac{8}{5 \sqrt{3}} \quad \therefore \quad \theta=\cos ^{-1} \frac{8}{5 \sqrt{3}}
\end{aligned}
$

Solution: (i) Given: Equation of one line is $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$
(It is standard form because coefficients of $x, y, z$ are unity each)
∴ Denominators 2, 5, – 3 are 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}
$

Given: Equation of second line is $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$
(It is also standard form)
∴ Denominators $-1,8,4$ are 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 the two given lines.
Recall that $\cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$
\begin{aligned}
& =\frac{2(-1)+5(8)+(-3)(4)}{\sqrt{4+25+9} \sqrt{1+64+16}} \quad \quad \quad \text { (From (i) and (ii)) } \\
& \Rightarrow \cos \theta=\frac{-2+40-12}{\sqrt{38} \sqrt{81}}=\frac{26}{9 \sqrt{38}} \Rightarrow \theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right) .
\end{aligned}
$
(ii) Given: Equation of one line is
$
\frac{x}{2}=\frac{y}{2}=\frac{z}{1}
$
(Standard Form)
∴ Denominators 2, 2, 1 are direction ratios of this line i.e., a vector along this line is
$
\overrightarrow{b_1}=(2,2,1)=2 \hat{i}+2 \hat{j}+\hat{k}
$

Given: Equation of second line is
$
\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}
$
(Standard Form)
∴ Denominators 4, 1, 8 are direction ratios of this line i.e., a vector along this line is
$
\overrightarrow{b_2}=(4,1,8)=4 \hat{i}+\hat{j}+8 \hat{k}
$

Let $\theta$ be the angle between the two lines.
Recall that $\cos \theta=\frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$
\begin{aligned}
& =\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} \\
& \therefore \quad \theta=\cos ^{-1} \frac{2}{3} .
\end{aligned}
$

Solution: Let us put the equations of these lines in standard form (i.e., making coeff. of $x, y, z$ unity in each of them)






The first line can be written as
$
-\frac{(x-1)}{3}=\frac{7(y-2)}{2 p}=\frac{z-3}{2} \quad \text { or } \quad \frac{x-1}{-3}=\frac{y-2}{\left(\frac{2 p}{7}\right)}=\frac{z-3}{2}
$
$\therefore \quad$ direction ratios of this line are $-3, \frac{2 p}{7}, 2=a_1, b_1, c_1$.
And the equation of 2 nd line can be written as
$
\frac{-7(x-1)}{3 p}=\frac{y-5}{1}=\frac{-(z-6)}{5} \text { or } \frac{x-1}{-\frac{3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}
$
∴ The direction ratios of 2nd line are $\frac{-3 p}{7}, 1,-5=a_2, b_2, c_2$.
∵ The two lines are perpendicular, therefore
$
\begin{aligned}
& a_1 a_2+b_1 b_2+c_1 c_2=0 \\
\Rightarrow & -3\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+2 \times(-5)=0 \\
\Rightarrow & \frac{9 p}{7}+\frac{2 p}{7}-10=0 \quad \Rightarrow \frac{11 p}{7}=10 \quad \Rightarrow p=\frac{70}{11} .
\end{aligned}
$

Solution: Given: Equation of one line is
$
\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}
$
(Standard form)
Direction ratios of this line are its denominators $7,-5,1$
$
=a_1, b_1, c_1\left(\Rightarrow \overrightarrow{b_1}=7 \hat{i}-5 \hat{j}+\hat{k}\right)
$

Given: Equation of second line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ (Standard form)
Direction ratios of this line are its denominators $1,2,3$
$
\begin{gathered}
\quad=a_2, b_2, c_2 \quad\left(\Rightarrow \quad \overrightarrow{b_2}=\hat{i}+2 \hat{j}+3 \hat{k}\right) \\
\overrightarrow{b_1} \cdot \overrightarrow{b_2}=a_1 a_2+b_1 b_2+c_1 c_2=7(1)+(-5)(2)+1(3) \\
=7-10+3=0
\end{gathered}
$
∴ The two given lines are perpendicular to each other.

Solution: Comparing the equations of the given lines with
$\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$, we have
$
\overrightarrow{a_1}=\hat{i}+2 \hat{j}+\hat{k}, \overrightarrow{b_1}=\hat{i}-\hat{j}+\hat{k}
$
and $\overrightarrow{a_2}=2 \hat{i}-\hat{j}-\hat{k}, \overrightarrow{b_2}=2 \hat{i}+\hat{j}+2 \hat{k}$
Recall that the S.D. between the two skew lines is given by
$
d=\frac{\left.\mid \overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times b_2\right) \mid}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}
$

Now $\overrightarrow{a_2}-\overrightarrow{a_1}=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k}$
$
\begin{aligned}
& \begin{aligned}
\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}-0 \hat{j}+3 \hat{k}
\end{aligned} \\
& \begin{aligned}
\therefore \quad\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right| & =\sqrt{(-3)^2+0^2+3^2}=\sqrt{18}=3 \sqrt{2} \\
\text { Next }\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)(-3)+(-3)(0)+(-2)(3)=-9
\end{aligned}
\end{aligned}
$

Substituting these values in eqn. (i),
$
\text { S.D. }(d)=\frac{|-9|}{3 \sqrt{2}}=\frac{9}{3 \sqrt{2}}=\frac{3}{\sqrt{2}}=\frac{3 \sqrt{2}}{2} .
$

Solution: Equation of one line is $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$
Comparing 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, z_1=-1 ; a_1=7, b_1=-6, c_1=1
$
$\therefore \quad$ vector form of this line is $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$
where $\overrightarrow{a_1}=\left(x_1, y_1, z_1\right)=(-1,-1,-1)=-\hat{i}-\hat{j}-\hat{k}$
and $\overrightarrow{b_1}=a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k}=7 \hat{i}-6 \hat{j}+\hat{k}$





Equation of second line is $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$
Comparing with $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$; we have
$
x_2=3, y_2=5, z_2=7 ; a_2=1, b_2=-2, c_2=1
$
$\therefore \quad$ vector form of this second line is $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$
where $\overrightarrow{a_2}=\left(x_2, y_2, z_2\right)=(3,5,7)=3 \hat{i}+5 \hat{j}+7 \hat{k}$
and $\overrightarrow{b_2}=a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}=\hat{i}-2 \hat{j}+\hat{k}$
recall that S.D. between two skew lines is given by
$
d=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times b_2\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}
$

Now $\overrightarrow{a_2}-\overrightarrow{a_1}=3 \hat{i}+5 \hat{j}+7 \hat{k}-(-\hat{i}-\hat{j}-\hat{k})$
$
\begin{aligned}
& =4 \hat{i}+6 \hat{j}+8 \hat{k} \\
& \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right|
\end{aligned}
$
$
\begin{aligned}
& =(-6+2) \hat{i}-(7-1) \hat{j}+(-14+6) \hat{k} \\
& =-4 \hat{i}-6 \hat{j}-8 \hat{k}
\end{aligned}
$
$
\begin{aligned}
\operatorname{again}\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) & =4(-4)+6(-6)+8(-8) \\
& =-16-36-64=-116
\end{aligned}
$

Substituting these values in eqn. (i),
$
\text { S.D. } \begin{aligned}
(d)=\frac{|-116|}{\sqrt{116}} & =\frac{116}{\sqrt{116}}=\sqrt{116} \\
& =\sqrt{4 \times 29}=2 \sqrt{29}
\end{aligned}
$

Solution: 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})=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}
$

Comparing, $\overrightarrow{a_1}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{b_1}=\hat{i}-3 \hat{j}+2 \hat{k}$
Equation of second line is
$
\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})=\overrightarrow{a_2}+\mu \overrightarrow{b_2}
$

Comparing $\overrightarrow{a_2}=4 \hat{i}+5 \hat{j}+6 \hat{k}$ and $\overrightarrow{b_2}=2 \hat{i}+3 \hat{j}+\hat{k}$
Recall that the length of the shortest distance between two skew lines is
$
\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|}
$

Now $\overrightarrow{a_2}-\overrightarrow{a_1}=4 \hat{i}+5 \hat{j}+6 \hat{k}-(\hat{i}+2 \hat{j}+3 \hat{k})$
$
\begin{aligned}
& =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} \\
& \text { Next } \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 2 \\
2 & 3 & 1
\end{array}\right|
\end{aligned}
$

Expanding along the first row,
$
\begin{aligned}
& \overrightarrow{b_1} \times \overrightarrow{b_2}=\hat{i}(-3-6)-\hat{j}(1-4)+\hat{k}(3+6)=-9 \hat{i}+3 \hat{j}+9 \hat{k} \\
& \therefore \quad\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=3(-9)+3(3)+3(9) \\
& \quad=-27+9+27=9 \\
& \text { and } \quad \begin{aligned}
\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{81+9+81} \\
=\sqrt{171}=\sqrt{9 \times 19}=3 \sqrt{19}
\end{aligned}
\end{aligned}
$

Substituting these values in (i),
length of shortest distance $=\frac{|9|}{3 \sqrt{19}}=\frac{9}{3 \sqrt{19}}=\frac{3}{\sqrt{19}}$.

Solution: The first line is $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$




$
\begin{aligned}
& =\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})=\overrightarrow{a_1}+t \overrightarrow{b_1} \\
& \text { Comparing } \quad \overrightarrow{a_1}=\hat{i}-2 \hat{j}+3 \hat{k}, \overrightarrow{b_1}=-\hat{i}+\hat{j}-2 \hat{k}
\end{aligned}
$

The second line is $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$
$
\begin{aligned}
& =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})=\overrightarrow{a_2}+s \overrightarrow{b_2}
\end{aligned}
$

Comparing $\overrightarrow{a_2}=\hat{i}-\hat{j}-\hat{k}, \overrightarrow{b_2}=\hat{i}+2 \hat{j}-2 \hat{k}$
The shortest distance between the two skew lines is given by
$
d=\frac{\left.\mid \overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \mid}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}
$

Now $\overrightarrow{a_2}-\overrightarrow{a_1}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\hat{j}-4 \hat{k}$
$
\begin{aligned}
& \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 1 & -2 \\
1 & 2 & -2
\end{array}\right| \\
& =(-2+4) \hat{i}-(2+2) \hat{j}+(-2-1) \hat{k}=2 \hat{i}-4 \hat{j}-3 \hat{k} \\
& \begin{aligned}
\therefore & \overrightarrow{b_1} \times \overrightarrow{b_2} \mid=\sqrt{2^2+(-4)^2+(-3)^2}=\sqrt{29} \\
\text { Next }\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)(2)+(1)(-4)+(-4)(-3)=8
\end{aligned}
\end{aligned}
$

Substituting these values in eqn. (i),
$
\text { S.D. }(d)=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}} .
$

Solution: Given: Direction ratios of one line are $a, b, c$
⇒ A vector along this line is $\overrightarrow{b_1}=a \hat{i}+b \hat{j}+c \hat{k}$
Given: Direction ratios of second line are $b-c, c-a$, $a-b$ ⇒ A vector along the second line is
$
\overrightarrow{b_2}=(b-c) \hat{i}+(c-a) \hat{j}+(a-b) \hat{k}
$

Let $\theta$ be the angle between the two lines.
Recall that $\cos \theta=\frac{\left|\overrightarrow{b_1} \cdot \overrightarrow{b_2}\right|}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
$
\begin{aligned}
& \quad=\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \\
& \text { or } \quad \cos \theta=\frac{a b-a c+b c-a b+a c-b c}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \\
& =\frac{0}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \\
& =0=\cos 90^{\circ} \quad \therefore \theta=90^{\circ} .
\end{aligned}
$

Solution: Recall that a unit vector along $x$ axis is $\hat{i}=\hat{i}+0 \hat{j}+0 \hat{k}$
∴ By definition, direction cosines of $x$-axis are coefficients of $\hat{i}, \hat{j}, \hat{k}$ in the unit vector i.e., $1,0,0=l, m, n$.
∴ Equation of the required line




passing through the origin $(0,0,0)$ and parallel to $x$-axis. (In fact this required line is $x$-axis itself)
is $\quad \frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0} \quad$ i.e., $\quad \frac{x}{1}=\frac{y}{0}=\frac{z}{0}$
Remark. Whenever it is not mentioned “find vector equation of the line (or plane)”, we should find cartesian equation only.
However vector equation of the required line in the above question is $\quad \vec{r}=\vec{a}+\lambda \vec{b}$
(Here $\vec{a}=\overrightarrow{0}$ and $\vec{b}=\hat{i}$ )
$
\text { i.e., } \vec{r}=\overrightarrow{0}+\lambda \hat{i} \Rightarrow \overrightarrow{\boldsymbol{r}}=\lambda \hat{\boldsymbol{i}} .
$

Solution: Given: Equation of one line is $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$
(It is standard form because coefficient of $x, y, z$ each is unity) Direction ratios of this line are its denominators
$
-3,2 k, 2=a_1, b_1, c_1
$
$\left(\Rightarrow\right.$ a vector along this line is $\left.\overrightarrow{b_1}=-3 \hat{i}+2 k \hat{j}+2 \hat{k}\right)$
Equation of second line is $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$ (Standard form)
Direction ratios of this line are its denominators
$
3 k, 1,-5=a_2, b_2, c_2
$
( ⇒ a vector along this line is $\overrightarrow{b_2}=3 k \hat{i}+\hat{j}-5 \hat{k}$ )
Because the lines are given to be perpendicular, therefore
$
\begin{array}{cc}
& \overrightarrow{b_1} \cdot \overrightarrow{b_2}\left(=a_1 a_2+b_1 b_2+c_1 c_2\right)=0 \\
\Rightarrow & (-3)(3 k)+(2 k)(1)+2(-5)=0 \\
\Rightarrow & -9 k+2 k-10=0 \\
\Rightarrow & -7 k=10 \Rightarrow k=\frac{-10}{7}
\end{array}
$

Solution: Given: vector equation of one line is
$
\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})
$

Comparing with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ we have
$
\overrightarrow{a_1}=6 \hat{i}+2 \hat{j}+2 \hat{k} \text { and } \overrightarrow{b_1}=\hat{i}-2 \hat{j}+2 \hat{k}
$

Given: Vector equation of second line is
$
\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})
$




Comparing with $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$, we have
$
\overrightarrow{a_2}=-4 \hat{i}-\hat{k} \quad \text { and } \quad \overrightarrow{b_2}=3 \hat{i}-2 \hat{j}-2 \hat{k}
$

Recall that the length of the shortest distance between two skew lines is
$
\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|}
$

Now $\overrightarrow{a_2}-\overrightarrow{a_1}=-4 \hat{i}-\hat{k}-(6 \hat{i}+2 \hat{j}+2 \hat{k})$
$
=-4 \hat{i}-\hat{k}-6 \hat{i}-2 \hat{j}-2 \hat{k}=-10 \hat{i}-2 \hat{j}-3 \hat{k}
$

Next $\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2\end{array}\right|$
Expanding along the first row
$
\begin{aligned}
& =\hat{i}(4+4)-\hat{j}(-2-6)+\hat{k}(-2+6)=8 \hat{i}+8 \hat{j}+4 \hat{k} \\
& \therefore \quad\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=(-10) 8+(-2) 8+(-3) 4 \\
& =-80-16-12=-108 \\
& \text { and } \quad \begin{aligned}
\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right| & =\sqrt{(8)^2+(8)^2+(4)^2} \\
= & \sqrt{64+64+16}=\sqrt{144}=12
\end{aligned}
\end{aligned}
$

Substituting these values in (i), length of shortest distance
$
=\frac{|-108|}{12}=\frac{108}{12}=9 .
$

Solution: Given: A point on the required line is $\mathrm{A}(1,2,-4)$.
∴ Position vector of point A is $\vec{a}=(1,2,-4)$
$
=\hat{i}+2 \hat{j}-4 \hat{k}
$
$
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}
$

Equations of the two given lines are
$
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}
$

(Standard Form)
$
\text { and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
$

(Standard Form)

∴ Direction ratios of first given line are its denominators $3,-16,7$ i.e., a vector along this line is $\overrightarrow{b_1}=3 \hat{i}-16 \hat{j}+7 \hat{k}$ and direction ratios of the second given line are also its denominators 3, 8, – 5
i.e., a vector along the second line is $\overrightarrow{b_2}=3 \hat{i}+8 \hat{j}-5 \hat{k}$
Let $\vec{b}$ be the vector along the required line perpendicular to the two given lines.
∴ By definition of cross-product $\vec{b}=\overrightarrow{b_1} \times \overrightarrow{b_2}$
$
\text { i.e., } \quad \vec{b}=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|
$

Expanding along the first row
$
=\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)
$




$
=24 \hat{i}+36 \hat{j}+72 \hat{k}, \vec{b}=12(2 \hat{i}+3 \hat{j}+6 \hat{k})
$
∴ Equation of the required line is $\vec{r}=\vec{a}+\lambda \vec{b}$
Substituting values of $\vec{a}$ and $\vec{b}$,
$
\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(12)(2 \hat{i}+3 \hat{j}+6 \hat{k})
$

Replacing $12 \lambda$ by $\lambda$,
$
\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) .
$