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 a1, b1, c1 and a2, b2, c2 are perpendicular, if a1a2+b1b2+c1c2=0

∴-3(3k)+2k×1+2(-5)=0
⇒-9k+2k-10=0
⇒7k=-10
⇒k=-10/7

Therefore, for k=-10/7, the given lines are perpendicular to each other.