Question

Find the position vector of point P such that OP is inclined to the X-axis at 45° and to the Y-axis at 60° and OP = 12 units.

Answer 1

Let l, m, n be the direction cosines of OP.

Since OP is inclined to X-axis at 45° and to Y-axis at 60°, we get

`l = cos45^@ = 1/(sqrt2) and m = cos60^@ = 1/2`

Now, l² + m² + n² = 1

∴ `(1/(sqrt2))^2 + (1/2)^2 + n^2 = 1`

∴ `1/2 + 1/4 + n^2 = 1`

∴ `n^2 = 1 - 3/4`

∴ `n^2 = 1/4`

∴ `n = pm1/2`

Let `barr` be the position vector of the point P w.r.t. the origin.

Then `barr = bar(OP) = |barr| hatr`, where `|barr|` = OP = 12

∴ `barr = 12(lhati + mhatj + nhatk) = 12(1/(sqrt2)hati + 1/2 hatj pm 1/2 hatk)` 

Hence, `barr = 6sqrt(2hati) + 6hatj pm 6hatk`.