Solution

Two lines with direction cosines, $l_1, m_1, n_1$ and $l_2, m_2, n_2$, 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, $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$, we obtain

$ \begin{aligned} l_1 l_2+m_1 m_2+n_1 n_2 & =\frac{12}{13} \times \frac{4}{13}+(\frac{-3}{13}) \times \frac{12}{13}+(\frac{-4}{13}) \times \frac{3}{13} \\ & =\frac{48}{169}-\frac{36}{169}-\frac{12}{169} \\ & =0 \end{aligned} $

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines, $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ and $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$, we obtain

$ \begin{aligned} l_1 l_2+m_1 m_2+n_1 n_2 & =\frac{4}{13} \times \frac{3}{13}+\frac{12}{13} \times(\frac{-4}{13})+\frac{3}{13} \times \frac{12}{13} \\ & =\frac{12}{169}-\frac{48}{169}+\frac{36}{169} \\ & =0 \end{aligned} $

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines, $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ and $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$, we obtain

$ \begin{aligned} l_1 l_2+m_1 m_2+n_1 n_2 & =(\frac{3}{13}) \times(\frac{12}{13})+(\frac{-4}{13}) \times(\frac{-3}{13})+(\frac{12}{13}) \times(\frac{-4}{13}) \\ & =\frac{36}{169}+\frac{12}{169}-\frac{48}{169} \\ & =0 \end{aligned} $

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.

Solution

Let $A B$ be the line joining the points, $(1,-1,2)$ and $(3,4,-2)$, and $C D$ be the line joining the points, $(0,3,2)$ and $(3,5,6)$.

The direction ratios, $a_1, b_1, c_1$, of $AB$ are $(3-1),(4-(-1))$, and $(-2-2)$ i.e., 2,5 , and -4 .

The direction ratios, $a_2, b_2, c_2$, of $C D$ are $(3-0),(5-3)$, and $(6-2)$ i.e., 3,2 , and 4.

$A B$ and $C D$ will be perpendicular to each other, if $a_1 a_2+b_1 b_2+c_1 c_2=0$

$a_1 a_2+b_1 b_2+c_1 c_2=2 \times 3+5 \times 2+(-4) \times 4$

$=6+10-16$

$=0$

Therefore, $A B$ and $C D$ are perpendicular to each other.

Solution

Let $A B$ be the line through the points, $(4,7,8)$ and $(2,3,4)$, and $C D$ be the line through the points, $(-1,-2,1)$ and $(1,2,5)$.

The directions ratios, $a_1, b_1, c_1$, of $A B$ are $(2-4),(3-7)$, and $(4-8)$ i.e., $-2,-4$, and -4 .

The direction ratios, $a_2, b_2, c_2$, of $C D$ are $(1-(-1)),(2-(-2))$, and (5 - 1) i.e., 2 , 4 , and 4. $AB$ will be parallel to $CD$, if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{a_1}{a_2}=\frac{-2}{2}=-1$

$\frac{b_1}{b_2}=\frac{-4}{4}=-1$

$\frac{c_1}{c_2}=\frac{-4}{4}=-1$

$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Thus, $A B$ is parallel to $C D$.

Solution

It is given that the line passes through the point $A(1,2,3)$. Therefore, the position vector through A is $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$

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

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 $\lambda$ is a constant.

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

This is the required equation of the line.

Solution

It is given that the line passes through the point with position vector $\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}$

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

It is known that a line through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is given by the equation, $\vec{r}=\vec{a}+\lambda \vec{b}$

$\Rightarrow \vec{r}=2 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+2 \hat{j}-\hat{k})$

This is the required equation of the line in vector form.

$\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$

$\Rightarrow x \hat{i}+y \hat{j}+z \hat{k}=(\lambda+2) \hat{i}+(2 \lambda-1) \hat{j}+(-\lambda+4) \hat{k}$

Eliminating $\lambda$, we obtain the Cartesian form equation as

$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$

This is the required equation of the given line in Cartesian form.

Solution

It is given that the line passes through the point $(-2,4,-5)$ and is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

The direction ratios of the line, $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$, are 3,5 , and 6 .

The required line is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

Therefore, its direction ratios are $3 k, 5 k$, and $6 k$, where $k \neq 0$

It is known that the equation of the line through the point $(x_1, y_1, z_1)$ and with direction

ratios, $a, b, c$, is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Therefore the equation of the required line is

$\frac{x+2}{3 k}=\frac{y-4}{5 k}=\frac{z+5}{6 k}$

$\Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k$

Solution

The Cartesian equation of the line is

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

The given line passes through the point $(5,-4,6)$. The position 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 , and 2 .

This means that the line is in the direction of vector, $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$

It is known that the line through position vector $\vec{a}$ and in the direction of the $\vec{b}$ is given by the equation, $\vec{r}=\vec{a}+\lambda \vec{b}, \lambda \in R$

$\Rightarrow \vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$

This is the required equation of the given line in vector form.

Solution

(i) Let $Q$ be the angle between the given lines.

The angle between the given pairs of lines is given by, $\cos Q=|\frac{ \vec{b} _1 \cdot \vec{b} _2}{| \vec{b} _1|| \vec{b} _2|}|$

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.

$ \begin{aligned} \therefore| \vec{b} _1| & =\sqrt{3^{2}+2^{2}+6^{2}}=7 \\ | \vec{b} _2| & =\sqrt{(1)^{2}+(2)^{2}+(2)^{2}}=3 \\ \vec{b} _1 \cdot \vec{b} _2 & =(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k}) \\ & =3 \times 1+2 \times 2+6 \times 2 \\ & =3+4+12 \\ & =19 \end{aligned} $

$\Rightarrow \cos Q=\frac{19}{7 \times 3}$

$\Rightarrow Q=\cos ^{-1}(\frac{19}{21})$

(ii) 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.

$ \begin{aligned} & \therefore| \vec{b} _1|=\sqrt{(1)^{2}+(-1)^{2}+(-2)^{2}}=\sqrt{6} \\ & | \vec{b} _2|=\sqrt{(3)^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{50}=5 \sqrt{2} \\ & \vec{b} _1 \cdot \vec{b} _2=(\hat{i}-\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}-4 \hat{k}) \\ & =1 \cdot 3-1(-5)-2(-4) \\ & =3+5+8 \\ & =16 \\ & \cos Q=|\frac{ \vec{b} _1 \cdot \vec{b} _2}{| \vec{b} _1|| \vec{b} _2|}| \\ & \Rightarrow \cos Q=\frac{16}{\sqrt{6} \cdot 5 \sqrt{2}}=\frac{16}{\sqrt{2} \cdot \sqrt{3} \cdot 5 \sqrt{2}}=\frac{16}{10 \sqrt{3}} \\ & \Rightarrow \cos Q=\frac{8}{5 \sqrt{3}} \\ & \Rightarrow Q=\cos ^{-1}(\frac{8}{5 \sqrt{3}}) \end{aligned} $

Solution

Let $ \vec{b} _1$ and $ \vec{b} _2$ be the vectors parallel to the pair of lines,

$ \begin{aligned} & \frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3} \text{ and } \frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4} \text{, respectively. } \\ & \therefore \vec{b} _1=2 \hat{i}+5 \hat{j}-3 \hat{k} \text{ and } \vec{b} _2=-\hat{i}+8 \hat{j}+4 \hat{k} \\ & | \vec{b} _1|=\sqrt{(2)^{2}+(5)^{2}+(-3)^{2}}=\sqrt{38} \\ & | \vec{b} _2|=\sqrt{(-1)^{2}+(8)^{2}+(4)^{2}}=\sqrt{81}=9 \\ & \vec{b} _1 \cdot \vec{b} _2=(2 \hat{i}+5 \hat{j}-3 \hat{k}) \cdot(-\hat{i}+8 \hat{j}+4 \hat{k}) \\ & =2(-1)+5 \times 8+(-3) \cdot 4 \\ & =-2+40-12 \\ & =26 \end{aligned} $

The angle, $Q$, between the given pair of lines is given by the relation,

$ \begin{aligned} & \cos Q=|\frac{ \vec{b} _1 \cdot \vec{b} _2}{| \vec{b} _1|| \vec{b} _2|}| \\ & \Rightarrow \cos Q=\frac{26}{9 \sqrt{38}} \\ & \Rightarrow Q=\cos ^{-1}(\frac{26}{9 \sqrt{38}}) \end{aligned} $

(ii) Let $ \vec{b} _1, \vec{b} _2$ be the vectors parallel to the given pair of lines, $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-5}{1}=\frac{z-3}{8}$, respectively.

$ \begin{aligned} & \vec{b} _1=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{b} _2=4 \hat{i}+\hat{j}+8 \hat{k} \\ & \therefore| \vec{b} _1|=\sqrt{(2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{9}=3 \\ & | \vec{b} _2|=\sqrt{4^{2}+1^{2}+8^{2}}=\sqrt{81}=9 \\ & \vec{b} _1 \cdot \vec{b} _2=(2 \hat{i}+2 \hat{j}+\hat{k}) \cdot(4 \hat{i}+\hat{j}+8 \hat{k}) \\ & \quad=2 \times 4+2 \times 1+1 \times 8 \\ & \quad=8+2+8 \\ & \quad=18 \end{aligned} $

If $Q$ is the angle between the given pair of lines, then $\cos Q=|\frac{ \vec{b} _1 \cdot \vec{b} _2}{| \vec{b} _1|| \vec{b} _2|}|$

$\Rightarrow \cos Q=\frac{18}{3 \times 9}=\frac{2}{3}$

$\Rightarrow Q=\cos ^{-1}(\frac{2}{3})$

Solution

The given equations can be written in the standard form as

$ \frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text{ and } \frac{x-1}{\frac{-3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5} $

The direction ratios of the lines are $-3, \frac{2 p}{7}, 2$ and $\frac{-3 p}{7}, 1,-5$ respectively.

Two lines with direction ratios, $a_1, b_1, c_1$ and $a_2, b_2, c_2$, are perpendicular to each other, if $a_1 a_2+b_1 b_2+c_1 c_2=0$ $\therefore(-3) \cdot(\frac{-3 p}{7})+(\frac{2 p}{7}) \cdot(1)+2 \cdot(-5)=0$

$\Rightarrow \frac{9 p}{7}+\frac{2 p}{7}=10$

$\Rightarrow 11 p=70$

$\Rightarrow p=\frac{70}{11}$

Thus, the value of $p$ is $\frac{70}{11}$.

Solution

The equations of the given lines are $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$

The direction ratios of the given lines are $7,-5,1$ and $1,2,3$ respectively.

Two lines with direction ratios, $a_1, b_1, c_1$ and $a_2, b_2, c_2$, are perpendicular to each other, if $a_1 a_2+b_1 b_2+c_1 c_2=0$

$\therefore 7 \times 1+(-5) \times 2+1 \times 3$

$=7-10+3$

$=0$

Therefore, the given lines are perpendicular to each other.

Solution

The equations of the given lines are

$\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$

$\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})$

It is known that the shortest distance between the lines, $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\mu \vec{b} _2$, is given by,

$d=|\frac{( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _2)}{|\overrightarrow{{}b_1} \times \vec{b} _2|}|$

Comparing the given equations, we obtain

$ \begin{aligned} & \vec{a} _1=\hat{i}+2 \hat{j}+\hat{k} \\ & \vec{b} _1=\hat{i}-\hat{j}+\hat{k} \\ & \vec{a} _2=2 \hat{i}-\hat{j}-\hat{k} \\ & \vec{b} _2=2 \hat{i}+\hat{j}+2 \hat{k} \\ & \vec{a} _2- \vec{a} _1=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k} \\ & \vec{b} _1 \times \vec{b} _2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \\ & \vec{b} _1 \times \vec{b} _2=(-2-1) \hat{i}-(2-2) \hat{j}+(1+2) \hat{k}=-3 \hat{i}+3 \hat{k} \\ & \Rightarrow| \vec{b} _1 \times \vec{b} _2|=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \end{aligned} $

Substituting all the values in equation (1), we obtain

$ \begin{aligned} & d=|\frac{(-3 \hat{i}+3 \hat{k}) \cdot(\hat{i}-3 \hat{j}-2 \hat{k})}{3 \sqrt{2}}| \\ & \Rightarrow d=|\frac{-3.1+3(-2)}{3 \sqrt{2}}| \\ & \Rightarrow d=|\frac{-9}{3 \sqrt{2}}| \\ & \Rightarrow d=\frac{3}{\sqrt{2}}=\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3 \sqrt{2}}{2} \end{aligned} $

Therefore, the shortest distance between the two lines is $\frac{3 \sqrt{2}}{2}$ units.

Solution

The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

It is known that the shortest distance between the two lines,

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$, is given by,

$d=\frac{ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{vmatrix} }{\sqrt{(b_1 c_2-b_2 c_1)^{2}+(c_1 a_2-c_2 a_1)^{2}+(a_1 b_2-a_2 b_1)^{2}}}$

Comparing the given equations, we obtain

$ \begin{aligned} & x_1=-1, y_1=-1, z_1=-1 \\ & a_1=7, \quad b_1=-6, c_1=1 \\ & x_2=3, \quad y_2=5, z_2=7 \\ & a_2=1, \quad b_2=-2, c_2=1 \\ & \text{ Then, } \begin{vmatrix} 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{vmatrix} = \begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \\ & =4(-6+2)-6(7-1)+8(-14+6) \\ & =-16-36-64 \\ & =-116 \\ & \Rightarrow \sqrt{(b_1 c_2-b_2 c_1)^{2}+(c_1 a_2-c_2 a_1)^{2}+(a_1 b_2-a_2 b_1)^{2}}=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}} \\ & =\sqrt{16+36+64} \\ & =\sqrt{116} \\ & =2 \sqrt{29} \end{aligned} $

Substituting all the values in equation (1), we obtain

$d=\frac{-116}{2 \sqrt{29}}=\frac{-58}{\sqrt{29}}=\frac{-2 \times 29}{\sqrt{29}}=-2 \sqrt{29}$

Since distance is always non-negative, the distance between the given lines is $2 \sqrt{29}$ units.

Solution

The given lines are $\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$

It is known that the shortest distance between the lines, $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\mu \vec{b} _2$, is given by, $d=|\frac{( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _2)}{| \vec{b} _1 \times \vec{b} _2|}|$

Comparing the given equations with $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\mu \vec{b} _2$, we obtain

$ \begin{aligned} & \vec{a} _1=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \vec{b} _1=\hat{i}-3 \hat{j}+2 \hat{k} \\ & \vec{a} _2=4 \hat{i}+5 \hat{j}+6 \hat{k} \\ & \vec{b} _2=2 \hat{i}+3 \hat{j}+\hat{k} \\ & \vec{a} _2- \vec{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} \\ & \vec{b} _1 \times \vec{b} _2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} =(-3-6) \hat{i}-(1-4) \hat{j}+(3+6) \hat{k}=-9 \hat{i}+3 \hat{j}+9 \hat{k} \\ & \Rightarrow| \vec{b} _1 \times \vec{b} _2|=\sqrt{(-9)^{2}+(3)^{2}+(9)^{2}}=\sqrt{81+9+81}=\sqrt{171}=3 \sqrt{19} \\ & ( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _1)=(-9 \hat{i}+3 \hat{j}+9 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+3 \hat{k}) \\ & =-9 \times 3+3 \times 3+9 \times 3 \\ & =9 \end{aligned} $

Substituting all the values in equation (1), we obtain

$d=|\frac{9}{3 \sqrt{19}}|=\frac{3}{\sqrt{19}}$

Therefore, the shortest distance between the two given lines is $\frac{3}{\sqrt{19}}$ units.

Solution

The given lines are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$

$\Rightarrow \vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})$

$\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$

$\Rightarrow \vec{r}=(\hat{i}-\hat{j}+\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})$

It is known that the shortest distance between the lines, $\vec{r}= \vec{a} _1+\lambda \vec{b} _1$ and $\vec{r}= \vec{a} _2+\mu \vec{b} _2$, is given by,

$d=|\frac{( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _2)}{| \vec{b} _1 \times \vec{b} _2|}|$

For the given equations,

$ \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}$

$ \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}$

$ \vec{b} _1 \times \vec{b} _2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2\end{vmatrix} =(-2+4) \hat{i}-(2+2) \hat{j}+(-2-1) \hat{k}=2 \hat{i}-4 \hat{j}-3 \hat{k}$

$\Rightarrow| \vec{b} _1 \times \vec{b} _2|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=\sqrt{29}$

$\therefore( \vec{b} _1 \times \vec{b} _2) \cdot( \vec{a} _2- \vec{a} _1)=(2 \hat{i}-4 \hat{j}-3 \hat{k}) \cdot(\hat{j}-4 \hat{k})=-4+12=8$

Substituting all the values in equation (3), we obtain

$d=|\frac{8}{\sqrt{29}}|=\frac{8}{\sqrt{29}}$

Therefore, the shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units.