The value of \(x\) for which the lines \(\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\) and \(\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular is
Answer & explanation
Correct answer: option 3
Lines are perpendicular there for \(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0\)
⇒ −3(3k)+2k×1+2(−5)=0
$ \Rightarrow 7k = -10$
$\Rightarrow k = \frac{-10}{7}$