Question

Find the value of $k$ if the pts $A\equiv (1,2,-1),B\equiv (4,-2,4)$ and $C\equiv (0,0,k)$ form a triangle right angled at $C$.

Answer 1

$DRS$ of

$BA\equiv (3.-4,5)$

$CA\equiv (1,2,-1-K)$

$CB\equiv (4,-2,4-K)$

since $CA\perp CB$

$\therefore \overrightarrow{CA}.\overrightarrow{CB}=0$

$\therefore \left(1\right)\left(4\right)+\left(2\right)(-2)+(-1-K)(4-K)=0$

$\Rightarrow 4-4+(K+1)(K-4)=0$

$\Rightarrow (K+1)(K-4)=0$

$\Rightarrow K=-1$ OR $K=4$