Discussion Forum

If P(6,10,10), Q(1,0,-5) , R (6,-10, $\lambda$)  are vertices of a triangle right angled at Q, then value of $\lambda$ is 

Answer:

Explanation:

In $\triangle$ PQR is right angled , at $\theta$

$\therefore$    PR²=PQ²+RQ²

 $\Rightarrow (6-6)^{2}+(-10-10)^{2}+(\lambda-10)^{2}+[(1-6)^{2}+(0-10)^{2}+(-5-10)^{2}]$

   = $[(1-6)^{2}+(0+10)^{2}+(-5-\lambda)^{2}]$

  $\Rightarrow$  $400+\lambda^{2}+100-20\lambda=(25+100+225)+(25+100+25+\lambda^{2}+10\lambda)$

 $\Rightarrow$  $\lambda^{2}+20\lambda+500=350+150+10\lambda+\lambda^{2}$

$\Rightarrow$  $-20\lambda=10\lambda\Rightarrow30\lambda=0$

$\Rightarrow$   $\lambda=0$