Question

The shortest distance between the lines $\frac{x}{m_1}=\frac{y}{1}=\frac{z-a}{0},\frac{x}{m_2}=\frac{y}{1}=\frac{z+a}{0}$ is

• A. $0$
• B. $\frac{a}{|m_1-m_2|}$
• C. $2a$
• D. $\frac{2a}{|m_1-m_2|}$

Answer 1

The correct option is C $2a$

Line $L_1$: $\frac{x}{m_1}=\frac{y}{1}=\frac{z-a}{0}$

Line $L_2$: $\frac{x}{m_2}=\frac{y}{1}=\frac{z+a}{0}$

Since,$L_1$ is passing through point $a_1=$(0,0,a) and the $L_2$ is passing through point $a_2=$(0,0,-a)

and the direction vector of line $L_1$ is $b_1=m_1\hat{i}+\hat{j}$.

The direction vector of line $L_2$ is $b_2=m_2\hat{i}+\hat{j}$

$\overrightarrow{b_1}\times \overrightarrow{b_2}=$($m_1\hat{i}+\hat{j}$)$\times$ ($m_2\hat{i}+\hat{j}$)

$=>(m_1-m_2)\hat{k}$--(I)

$\overrightarrow{a_2}-\overrightarrow{a_1}=(0,0,-2a)$--(II)

the shortest distance between the lines is $d=\left|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$--(III)

put the value of (I) and (II) in (III) ,we get:

$d=\left|\frac{(-2a\hat{k})\cdot \left(\right(m_1-m_2\left)\hat{k}\right)}{(m_1-m_2)}\right|$

$d=2a$ unit

Hence ,Option C is correct.