vecctor `vec(OA)=hati+2hatj+2hatk` turns through a right angle passing through the positive x-axis on the way. Show that the vector in its new postion is `(4hati-hatj-hatk)/sqrt2`

Explanation

Let the new vector be `vec(OB)= xhati+yhatj=zhatk`
According to the given condition, we have
`|vec(OB)|=|vec(OA)|=3 Rightarrowx^(2)+y^(2)+z^(2)=9`
`vec(OA)vec(OB)Rightarrowx+2y+2z=0`
Since with turing `vec(OA)`, it passes through the postivie x-axis on the way, vectors `vec(OA),vec(OB)and lambdahati` coplanar . thus,
`|{:(x,y,z),(1,2,2),(lambda,0,0):}|=0`
or y-z=0
solving (i) (ii) and (iii) for x,y and z. we have x-4y=-4z
`Rightarrow 16y^(2)+y^(2)+y^(2)=9`
`Rightarrowy=+-1/sqrt2`
`Rightarrow vec(OB) = +-(4/sqrt2hati-1/sqrt2hatj-1/sqrt2hatk)`
since angle between `vec(OB) and hati` is acute, `vec(OB)=4/sqrt2hati-1/sqrt2hatj-1/sqrt2hatk`