If the line $\frac{-x+1}{3}=\frac{-y-2}{-2k}=\frac{z+3}{2}$ and $\frac{-1+x}{3k}=\frac{-1+ y}{1}=\frac{-z+6}{5}$ are perpendicular, then the value of $k$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $-\frac{10}{7}$
First line:
$-\frac{x+1}{3}=\frac{-y-2}{-2k}=\frac{z+3}{2}=t$
$x=-3t-1,\quad y=2kt-2,\quad z=2t-3$
Direction ratios: $(-3,\;2k,\;2)$
Second line:
$\frac{-1+x}{3k}=\frac{-1+y}{1}=\frac{-z+6}{5}=s$
$x=3ks+1,\quad y=s+1,\quad z=-5s+6$
Direction ratios: $(3k,\;1,\;-5)$
Condition for perpendicularity:
$(-3)(3k)+(2k)(1)+(2)(-5)=0$
$-9k+2k-10=0$
$-7k-10=0$
$k=-\frac{10}{7}$