If $\to a, \to b, \to c$ are three mutually perpendicular vectors, then the vector which is equally inclined to these vectors is?

\to a + \to b + \to c

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\frac{\to a}{| \to a |} + \frac{\to b}{| \to b |} + \frac{\to c}{| \to c |}

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\frac{\to a}{| \to a |^{2}} + \frac{\to b}{| \to b |^{2}} + \frac{\to c}{| \to c |^{2}}

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| \to a | \to a - | \to b | \to b + | \to c | \to c

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Solution

The correct option is C $\to a + \to b + \to c$
Atp,
$\to a . \to b + \to a . \to c + \to c . \to b = 0$
$| \to c + \to a + \to b |^{2} = | a |^{2} + | b |^{2} + | c |^{2}$
$\to a . (\to a + \to b + \to c) = | \to a |^{2}$
$θ = {cos}^{- 1} \frac{| \to a |^{2}}{| \to a | \times \sqrt{| \to a |^{2} + | \to b |^{2} + | \to c |^{2}}} = {cos}^{- 1} \frac{| \to a |}{\sqrt{| \to a |^{2} + | \to b |^{2} + | \to c |^{2}}}$
Similarily,for $\to b . | \to c + \to a + \to b |, \to c . | \to c + \to a + \to b |$
It is equally inclined if $\to a, \to b, \to c$ are of equal magnitude.

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