Question

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate 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

The correct option is D 2

Here, $d_1=dcos({90}^{\circ }-\alpha )$
$d_2=dcos({90}^{\circ }-\beta )$
and $d_3=dcos({90}^{\circ }-\gamma )$
$\therefore d_1=dsin\alpha$
$d_2=dsin\beta$
and $d_3=dsin\gamma$
$\therefore d_1^2+d_2^2+d_3^2=k{d}^2$
$\Rightarrow d^2(si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma )=k(d{)}^2$
$\therefore k=2$