Question

If `vec(a)=2i+2j+2k`, `vec(b)=-i+2j+k`, `vec(c)=3i+j` are such that `vec(a)+ jvec(b)` is perpendicular to `vec(c)`,find the value of `j`?

Answer 1

`j=8`

Explanation:

`costheta=((a+jb).c)/(abs(a+jb)abs(c))`

However, `theta=90`, so `cos90=0`

`(a+jb).c=0`

`a+jb=((2),(2),(2))+j((-1),(2),(1))=((2-j),(2+2j),(2+j))`

`c=((3),(1),(0))`

`(a+jb).c=3(2-j)+2+2j=6-3j+2+2j=8-j=0`

`j=8`