Prove that $(a + b) . (a + b) = | a |^{2} + | b |^{2}$ , if and only if a, b are perpendicular, given $a \neq 0$ , $b \neq 0$

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Solution

$(a + b) . (a + b) = | a |^{2} + | b |^{2} \Rightarrow a . (a + b) + b . (a + b) = | a |^{2} + | b |^{2} \Rightarrow a . a + a . b + b . a + b . b = | a |^{2} + | b |^{2} \Rightarrow | a |^{2} + 2 a . b + | b |^{2} = | a |^{2} + | b |^{2}$
$2 a . b = 0$ $[∵ a . b = b . a]$
(scalar product is commutative)
$\Rightarrow a . b = 0$
$∴$ a and b are perpendicular

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