Question

If the line, $\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+\lambda }{-2}$ lies on the plane $2x-4y+3z=2$,

then the shortest distance between this line and the line $\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}$ is

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

Answer: (A)

$0$

$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+\lambda }{-2}=k$

Let P be any point on the line,

$P=(k+3,-k-2,-2k-\lambda )$

P lies on the plane $2x-4y+3z=2$

$2(k+3)-4(-k-2)+3(-2k-\lambda )=2$

$2k+6+4k+8-6k-3\lambda =2$

$14-3\lambda =2$

$3\lambda =12$

$\lambda =4$

$\therefore$ Line $1$ is $\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+4}{-2}$

Another line is $\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}$(line $2$)
Shortest distance be d.

$d=\frac{\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{\sum (a_1{b}_2-a_2{b}_1{)}^2}}$ where $a_1,b_1,c_1$ are d.r's of lines

$\to \left|\begin{array}{ccc}3-1& -2-0& -4-0\\ 1& -1& -2\\ 12& 9& 4\end{array}\right|$

$=\left|\begin{array}{ccc}2& -2& -4\\ 1& -1& -2\\ 12& 9& 4\end{array}\right|=|2\left(14\right)+2\left(28\right)-4\left(21\right)|=|28+56-84|=0$

$\therefore d=0$

$\Rightarrow d=0$.