Question

Prove that `( -> a+ -> b)dot( -> a+ -> c)| -> a|^2+| -> b|^2` , if and only if ` -> a , -> b` are perpendicular, given ` -> a!= ->0, -> b!= ->0`

Answer 1

`(vec(a) + vec(b)).(vec(a) + vec(b)) = |vec(a)|^(2)+|vec(b)|^(2)`
`implies |vec(a)|^(2) + |vec(b)|^(2) + 2vec(a).vec(b) = |vec(a)|^(2)+|vec(b)|^(2)`
`implies 2vec(a).vec(b) = 0implies vec(a).vec(b) = 0`
`implies vec(a)` and `vec(b)` are mutually perpendicular.
Hence Proved.