Question

If $l$, $m$, $n$ are the direction cosines of the line of shortest distance between the lines
$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ then

• A. $3l-15m+9n=0$
• B. $2l-7m+5n=0$
• C. $l=m=n=1/\sqrt{3}$
• D. $2l+m-3n=0$

Answer 1

Answer: (B), (C), (D)

$2l+m-3n=0$
$2l-7m+5n=0$
$l=m=n=1/\sqrt{3}$

From the given equations of the straight lines,

$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5},\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$

the given equations:

$\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ = $\lambda$ ....(1)

$\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ = $\mu$ .......(2)

General equations:

$\frac{x-a_1}{l_1}=\frac{y-b_1}{m_1}=\frac{z-c_1}{n_1}$ = $\lambda$ ....(3)

$\frac{x-a_2}{l_2}=\frac{y-m_2}{l_2}=\frac{z-n_2}{n_2}$ = $\mu$ .......(4)

$\Rightarrow$ If l, m, n are the direction cosines of the line of shortest distance between the given lines are

By comparing the equation (1) and (3):

$\Rightarrow$ $2l-7m+5n=0$.

By comparing the equation (2) and (4):

$\Rightarrow$ $2l+m-3n=0$

by solving the equation (1) and (2) we will get :

$\Rightarrow$ $l=m=n=$ $\frac{1}{\sqrt{3}}$.

Hence this problem has 3 answers.