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

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Solution

The correct option is D 2
Here, $d_{1} = d c o s (90^{\circ} - α)$
$d_{2} = d c o s (90^{\circ} - β)$
and $d_{3} = d c o s (90^{\circ} - γ)$
$∴ d_{1} = d s i n α$
$d_{2} = d s i n β$
and $d_{3} = d s i n γ$
$∴ d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = k d^{2}$
$\Rightarrow d^{2} (s i n^{2} α + s i n^{2} β + s i n^{2} γ) = k (d)^{2}$
$∴ k = 2$

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