Question

If `veca = hati + hatj + hatk` and `vecb = hati + 2hatj + 3hatk` then find a unit vector perpendicular to both `veca + vecb` and `veca - vecb`.

Answer 1

`veca = hati + hatj + hatk` and `vecb = hati + 2hatj + 3hatk`

Now `veca + vecb = (hati + hatj + hatk) + (hati + 2hatj + 3hatk)`

= `2hati + 3hatj + 4hatk`

and `veca - vecb = (hati + hatj + hatk) - (hati + 2hatj + 3hatk)`

= `-hatj - 2hatk`

So `(veca +  vecb) xx (veca - vecb)`

= `|(hati, hatj, hatk),(2, 3, 4),(0, -1, -2)|`

= `hati(-6 + 4) - hatj(-4) + hatk(-2)`

= `-2hati + 4hatj - 2hatk`

∴ Unit vector perpendicular to `(veca + vecb)` and `(veca - vecb)`.

= `(-2hati + 4hatj - 2hatk)/sqrt(4 + 16 + 4)`

= `(-2hati + 4hatj - 2hatk)/(2sqrt(6))`

= `(-hati + 2hatj - hatk)/sqrt(6)`

Therefore, unit vector perpendicular to `(veca + vecb)` and `(veca - vecb)` is `1/sqrt(6) (-hati + 2hatj - hatk)`.