Question

If the lines $\frac{2x-1}{2}=\frac{3-y}{1}=\frac{z-1}{3}and \frac{x+3}{2}=\frac{z+1}{p}=\frac{y+2}{5}$ are perpendicular to each other, then $p$ is equal to

• A. $1$
• B. $-1$
• C. $10$
• D. $-\frac{7}{5}$
• E. $-19$

Answer 1

Answer: (A)

Given lines are $\frac{2 x-1}{2}=\frac{3-y}{1}=\frac{z-1}{3}$
$\Rightarrow \, \frac{x-1 / 2}{1}=\frac{y-3}{-1}=\frac{z-1}{3}\,\,\,\,\,\,\dots(i)$
and $ \frac{x+3}{2}=\frac{z+1}{p}=\frac{y+2}{5}\,\,\,\,\,\,\,\dots(ii)$
Since, both lines are perpendicular.
$\therefore \, (1)(2)+(-1)(5)+(3)(p)=0$
(by perpendicularity condition)
$\Rightarrow \,2-5+3 p=0$
$ \Rightarrow \,3 p=3$
$\therefore \, p=1$