A non vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj` and thepane determined by the vectors `hati-hatj,hati+hatk` then angle between `veca and hati-2hatj+2hatk` is = (A) `pi/2` (B) `pi/3` (C) `pi/6` (D) `pi/4`

Answer

`pi//4 or 3pi//4`

Explanation

A vector normal to the plane containing vectors ` hati and hati + hatz` is
`vecp=|{:(hati,hatj,hatk),(1,0,0),(1,1,0):}|=hatk`
A vector normal to the plane containing vectors ` hati- hatj, hati + hatk` is
`vecq=|{:(hati,hatj,hatk),(1,-1,0),(1,0,1):}|=-hati-hatj+hatk`
vector `veca` is parllel to vector `vecp xx vecq`
`vecp xx vecq=|{:(hati,hatj,hatk),(0,0,1),(-1,-1,1):}|=hati-hatj`
Therefore, a vector in direction of `veca` is `hati - hatj`
Now `theta` is the angle between `hata and hati - 2hatj + 2hatk` then
`cos theta=+-(1.1+(-1).(-2))/(sqrt(1+1)sqrt(1+4+4))=+-3/(sqrt(2).3)`
`Rightarrow +-1/sqrt2 Rightarrow theta=pi/4or (3pi)/4`