Find the position vector of a point `A` in space such that ` vec O A` is inclined at `60^0toO X` and at `45^0toO Ya n d| vec O A|=10u n i t sdot`

Explanation

since, `vec{{OA}}` is inclined at `60^{circ}` to `{OX}` and at `45^{circ}` to `{OY}`.

Let `overline{{OA}}` make angle `alpha` with `{OZ}`.

`therefore cos ^{2} 60^{circ}+cos ^{2} 45^{circ}+cos ^{2} alpha=1`


`Rightarrow(frac{1}{2})^{2}+(frac{1}{sqrt{2}})^{2}+cos ^{2} alpha=1[because 1^{2}+m^{2}+n^{2}=1]`


`Rightarrow frac{1}{4}+frac{1}{2}+cos ^{2} alpha=1 `

`Rightarrow cos ^{2} alpha=1-(frac{1}{2}+frac{1}{4}) `

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