Show that $| a | b + | b | a$ is perpendicular to $| a | b - | b | a$ for any two non-zero vectors a and b.

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Solution

Let $p = | a | b + | b | a$ and $q = | a | b - | b | a$

Then, $p . q = (| a | b + | b | a) . (| a | b - | b | a)$

$= | a |^{2} (b . b) - | a | | b | (b . a) + | b | | a | (a . b) - | b |^{2} (a . a)$

$= | a |^{2} | b |^{2} - | a | | b | (a . b) + | a | | b | (a . b) - | b |^{2} | a |^{2} = 0$

$\Rightarrow p ⊥ q$ ( $∵$ if c.d = 0 $\Rightarrow$ c is perpendicular to d)

Hence, |a| b + |b| a and |a| b - |b| a are perpendicular to each other for any two non-zero vectors a and b.

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