Question

The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are d₁, d₂, d₃. Then $d_{1}^{2} + d_{2}^{2} + d_{3}^{2}$= Kd² where K is –

• A. 1
• B. 5
• C. 3
• D. 2

Answer 1

Answer: (D)

2

Answer 2

Here d₁ = d cos (90° – a), d₂ = d cos (90° – b)

and d₃ = d cos (90° – g)

d₁ = d sin a, d₂ = d sin b, d₃ = d sin g

Ž $d_{1}^{2} + d_{2}^{2} + d_{3}^{2}$ = kd²

Ž d² (sin² a + sin² b + sin² g) = kd²

\ k = 2.