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^(@)` to `OX` at `45^(@)` to `OY`. Let `vec(OA)` makes angle `alpha` with `OZ`.
`:. cos^(2)60^(@)+cos^(2) 45^(@)+cos^(2)alpha=1`
`implies(1/2)^(2)+(1/(sqrt(2)))^(2)+cos^(2)alpha=1 [ :' l^(2)+m^(2)+n^(2)=1]`
`= 1/4+1/2+cos^(2)alpha=1`
`implies cos^(2)alpha=1-(1/2+1/4)`
`implies cos^(2)alpha=1-(6/8)`
`implies cos^(2) alpha=1/4`
`implies cos alpha=1/2=cos60^(@)`
`:. alpha=60^(@)`
`:. vec(OA)=|vec(OA)|(1/2hati+1/(sqrt(2))hatj+1/2hatk)`
`=10(1/2hati+1/(sqrt(2))hatj+1/2hatk) [ :' |vec(OA)|=10]`
`=5hati+5sqrt(2)hatj+5hatk`