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