Question

If the line $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular to each other then $k=$

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

Answer 1

The correct option is D $\frac{-10}{7}$

Given lines are $\perp$

$\Rightarrow L$ b/w their direction cosines is ${90}^o$.

$\Rightarrow .(-3,2k,2).(3k,1,5)=0$

$-9k+2k-10=0$

$-7k=10$

$k=-10/7$