Question

The distance between two points $P$ and $Q$ is $d$ and the length of their projections of $P Q$ on the co-ordinate planes are $d_{1}, d_{2}, d_{3}$. Then $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=k d^{2}$ where ' $k$ ' is

• A. 1
• B. 5
• C. 3
• D. 2

Answer 1

Answer: (D)

Here, $d_{1}=d \cos \left(90^{\circ}-\alpha\right)$ ;
$ d_{2}=d \cos \left(90^{\circ}-\beta\right)$
$d_{3}=d \cos \left(90^{\circ}-\gamma\right)$
$\Rightarrow d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=d^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)$
$\Rightarrow d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=2 d^{2} $;
$ \therefore k=2$