Question

$P(1,1,1)$ and $Q(\lambda ,\lambda ,\lambda )$ are two points in the space such that $PQ=\sqrt{27},$ the value of $\lambda$ can be

• A. $-4$
• B. $-2$
• C. $2$
• D. $0$

Answer 1

The correct option is B $-2$

${PQ}^2={(\lambda -1)}^2+{(\lambda -1)}^2+{(\lambda -1)}^2$

$\Rightarrow 3{(\lambda -1)}^2=27$

$\Rightarrow {(\lambda -1)}^2=9$

$\Rightarrow \lambda -1=\pm 3$

$\Rightarrow \lambda =1\pm 3$

$\Rightarrow \lambda =1+3,$or $1-3$

$\therefore \lambda =4,$ or $-2$