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Question
If `veca` and `vecb` are two vectors perpendicular to each other, prove that `(veca + vecb)^2 = (veca - vecb)^2`
Solution
`veca` and `vecb` are perpendicular to each other.
∴ `veca*vecb = vecb*veca` = 0 ...(i)
LHS = `(veca + vecb)^2`
= `(veca + vecb)*(veca + vecb)`
= `veca*(veca + vecb) + vecb(veca + vecb)`
= `veca*veca + veca*vecb + vecb*veca + vecb*vecb`
= `veca*veca + 0 + 0 + vecb*vecb` ....[By (1)]
= `|veca|^2 + |vecb|^2`
RHS = `(veca - vecb)^2`
= `(veca - vecb)*(veca - vecb)`
= `veca*(veca - vecb) + vecb(veca - vecb)`
= `veca*veca - veca*vecb - vecb*veca + vecb*vecb`
= `veca*veca + vecb*vecb` ....[By (i)]
= `|veca|^2 + |vecb|^2`
∴ LHS = RHS
Hence, `(veca + vecb)^2 = (veca - vecb)^2`
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