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 $\Delta P Q R$ 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) $
$\Rightarrow \lambda^{2}-20 \lambda+500=350+150+10 \lambda+\lambda^{2}$
$\Rightarrow -20 \lambda=10 \lambda \Rightarrow 30 \lambda=0$
$\Rightarrow \lambda=0$