Show that vectors $∣ ∣ ¯ ¯ ¯ a ∣ ∣ ∣ ∣ ¯ ¯ ¯ a + ∣ ∣ ¯ ¯ ¯ a ∣ ∣ ¯ ¯ b$ and $| b | ¯ ¯ ¯ a - | a ∣ ∣ ¯ ¯ b$ are orthogonal.

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Solution

Given that, $∣ ∣ \to a ∣ ∣ \to b + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a$ and $∣ ∣ \to a ∣ ∣ \to b - ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a$

Now ,

$(∣ ∣ \to a ∣ ∣ \to b + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a) .$ $(∣ ∣ \to a ∣ ∣ \to b - ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a)$ = $(∣ ∣ \to a ∣ ∣ \to b + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a) .$ $(∣ ∣ \to a ∣ ∣ \to b + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a) .$

$= {∣ ∣ \to a ∣ ∣}^{2} {\to b}^{2} - ∣ ∣ \to a ∣ ∣ \to b ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a ∣ ∣ \to a ∣ ∣ \to b - {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} {\to a}^{2}$

$∵ \to x = {∣ ∣ \to x ∣ ∣}^{2}$

$∴ {∣ ∣ \to a ∣ ∣}^{2} {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} - ∣ ∣ \to a ∣ ∣ \to b ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a ∣ ∣ \to a ∣ ∣ \to b - {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} {∣ ∣ \to a ∣ ∣}^{2}$

$= {∣ ∣ \to a ∣ ∣}^{2} {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} - ∣ ∣ \to a ∣ ∣ \to b ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a ∣ ∣ \to a ∣ ∣ \to b - {∣ ∣ \to a ∣ ∣}^{2} {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2}$

$= - ∣ ∣ \to a ∣ ∣ \to b ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a + ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a ∣ ∣ \to a ∣ ∣ \to b$

$= - ∣ ∣ \to a ∣ ∣ \to b ∣ ∣ ∣ \to b ∣ ∣ ∣ \to a + ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to a ∣ ∣ \to a \to b$

$= 0$

hence, both given vector are orthogonal

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