Question

Show that `|veca|vecb+|vecb|veca`  is perpendicular to `|veca|vecb-|vecb|veca,` for any two nonzero vectors `veca and vecb`.

Answer 1

We have, `|veca| vecb + |vecb| veca, |veca| vecb - |vecb| veca` is perpendicular to,

If `[|veca| vecb + |vecb| veca] xx [|veca| vecb - |vecb| veca] = 0`

= `|veca| vecb xx |veca| vecb - |veca| vecb| vecb| veca + |vecb| veca| vecb| xx veca - |vecb|^2 veca xx veca`

= `|veca|^2vecb xx vecb - veca||vecb|veca xx vecb + |vecb||veca||veca xx vecb - |vecb|^2|veca|^2`

= `|veca|^2 |vecb|^2  - |veca|^2 |vecb|^2`

= 0

These given circles are perpendicular to each other.

Hence proved.