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Question
Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.
Solution
Let a = 60°, b = 45° and the angle inclined to OZ axis be g
We know that cos2a + cos2b + cos2ϒ = 1
⇒ cos2 60° + cos2 45° + cos2ϒ = 1
⇒ `(1/2)^2 + (1/sqrt(2))^2 + cos^2ϒ` = 1
⇒ `1/4 + 1/2 + cos^2ϒ` = 1
⇒ `3/4 + cos^2ϒ` = 1
⇒ cos2ϒ = `1 - 3/4 = 1/4`
∴ cos ϒ = `+- 1/2`
⇒ cos ϒ = `1/2` .....(Rejecting cos ϒ = `- 1/2`, since ϒ < 90°)
∴ `vec"OA" = |vec"OA"| (1/2hat"i" + 1/sqrt(2)hat"j" + 1/2hat"k")`
= `10(1/2hat"i" + 1/sqrt(2)hat"j" + 1/2hat"k")`
= `5hat"i" + 5sqrt(2)hat"j" + 5hat"k"`
Hence, the position vector of A is `(5hat"i" + 5sqrt(2)hat"j" + 5hat"k")`.
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