Question

Find the position vector of a point A in space such that \(\vec {OA}\) is inclined at 60° to \(\vec {OX}\) and at 45° to \(\vec {OY}\) and \(\vec {OA}\) = 10 units.

Answer 1

We know that, cos² α + cos² β + cos² γ = 1

cos² 60^o + cos² 45^o + cos² γ = 1

(1/2)² + (1/√2)² + cos² γ = 1

¼ + ½ + cos² γ = 1

cos² γ = 1 – ¾ = ¼

So, cos γ = ± ½ ⇒ cos γ = ½ [Rejecting cos γ = – ½, as γ < 90^o]

Now,