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

To show | a | b +| b | a is perpendicular to | a | b -| b | a for a 0 and b 0, we need to show Dot product of | a | b +| b | a and | a | b -| b | a is zero. (| a | b +| b | a ) (| a | b -| b | a ) = (|a| b) (|a| b)-(|a| b) (|b| a)+(|b| a) (|a| b)-(|b| a) (|b| a) = | a |^ 2 b b -| b |^ 2 a a = | a |^ 2 | b |^ 2 -| b |^ 2 | a |^ 2 [Since, a b = b a ] = 0 | a | b +| b | a is perpendicular to | a | b -| b | a

Show that |a| b+|b| a is perpendicular to |a| b-|b| a for any two…

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Question

Show that $|\vec{a}| \vec{b}+|\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b}-|\vec{b}| \vec{a}$ for any two nonzero vectors $\vec a$ and $\vec b$.

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Answer

To show $|\vec{\mathrm{a}}| \vec{\mathrm{b}}+|\vec{\mathrm{b}}| \vec{\mathrm{a}}$ is perpendicular to $|\vec{\mathrm{a}}| \vec{\mathrm{b}}-|\vec{\mathrm{b}}| \vec{\mathrm{a}}$ for $\vec a \neq0$ and $\vec b \neq 0$,
we need to show Dot product of $|\vec{\mathrm{a}}| \vec{\mathrm{b}}+|\vec{\mathrm{b}}| \vec{\mathrm{a}}$ and $|\vec{\mathrm{a}}| \vec{\mathrm{b}}-|\vec{\mathrm{b}}| \vec{\mathrm{a}}$ is zero.
$(|\vec{\mathrm{a}}| \vec{\mathrm{b}}+|\vec{\mathrm{b}}| \vec{\mathrm{a}}) \cdot(|\vec{\mathrm{a}}| \vec{\mathrm{b}}-|\vec{\mathrm{b}}| \vec{\mathrm{a}})$
= $(|\vec{a}| \vec{b}) \cdot(|\vec{a}| \vec{b})-(|\vec{a}| \vec{b}) \cdot(|\vec{b}| \vec{a})+(|\vec{b}| \vec{a}) \cdot(|\vec{a}| \vec{b})-(|\vec{b}| \vec{a}) \cdot(|\vec{b}| \vec{a})$
= $|\vec{\mathrm{a}}|^{2} \vec{\mathrm{b}} \cdot \vec{\mathrm{b}}-|\vec{\mathrm{b}}|^{2} \vec{\mathrm{a}} \cdot \vec{\mathrm{a}}$
= $|\vec{\mathrm{a}}|^{2}|\vec{\mathrm{b}}|^{2}-|\vec{\mathrm{b}}|^{2}|\vec{\mathrm{a}}|^{2}$ [Since, $\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}$]
= 0
$\Rightarrow$ $|\vec{\mathrm{a}}| \vec{\mathrm{b}}+|\vec{\mathrm{b}}| \vec{\mathrm{a}}$ is perpendicular to $|\vec{\mathrm{a}}| \vec{\mathrm{b}}-|\vec{\mathrm{b}}| \vec{\mathrm{a}}$

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