Question

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}and\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ perpendicular, then find the value of k.

Answer 1

Direction ratio of the given two lines are respectively (-3, 2k, 2) and (3k, 1, -5).
It is known that two lines with direction ratios, $a_1,b_1,c_{1}and{a}_2,b_2,c_2$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\therefore (-3)\times \left(3k\right)+\left(2k\right)\times 1+2\times (-5)=0\therefore -9k+2k-10=0\Rightarrow 7k=-10\Rightarrow k=\frac{-10}{7}$
Therefore, for $k=\frac{-10}{7}$, the given lines are perpendicular to each other.