Question

If lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$
and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$
are mutually perpendicular, then $k$ is equal to

• A. $-\frac{10}{7}$
• B. $-\frac{7}{10}$
• C. $-10$
• D. $-7$

Answer 1

Answer: (A)

Lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$
are perpendicular if $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$
$\Rightarrow -3\left(3k\right)+2k+2\left(-5\right)=0$
$\Rightarrow k-=\frac{10}{7}$