Question

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

Answer 1

The direction of ratios of 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 $-3,2k,2$ and $3k,1,-5$ respectively.
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 $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\therefore$ $-3\left(3k\right)+2k\times 1+2(-5)=0$
$\Rightarrow$ $-9k+2k-10=0$
$\Rightarrow$ $7k=-10$
$\Rightarrow$ $k=\frac{-10}{7}$
Therefore for $k=-\frac{10}{7}$, the given lines are perpendicular to each other.