Question

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 ...

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

Answer: (A)

In $\Delta PQR$ is right angled, at $\theta$ $\therefore$
$\therefore P{R}^2=P{Q}^2+R{Q}^2$
$\Rightarrow (6-6{)}^2+(-10-10{)}^2+(\lambda -10{)}^2+\left[(1-6{)}^2+(0-10{)}^2+(-5-10{)}^2\right]$
$=\left[(1-6{)}^2+(0+10{)}^2+(-5-\lambda {)}^2\right]$
$\Rightarrow 400+{\lambda }^2+100-20\lambda =(25+100+225)+\left(25+100+25+{\lambda }^2+10\lambda \right)$
$\Rightarrow {\lambda }^2-20\lambda +500=350+150+10\lambda +{\lambda }^2$
$\Rightarrow -20\lambda =10\lambda \Rightarrow 30\lambda =0$
$\Rightarrow \lambda =0$