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

The correct option is A $1$

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