If $\to a, \to b, \to c$ are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors $\to a$ and $\to a + \to b + \to c$ is

\frac{π}{3}

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\frac{π}{6}

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{cos}^{- 1} (\frac{1}{\sqrt{3}})

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\frac{π}{2}

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Solution

The correct option is B ${cos}^{- 1} (\frac{1}{\sqrt{3}})$
Since $\to a$ , $\to b$ and $\to c$ are mutually perpendicular so,
$\to a \cdot \to b = \to b \cdot \to c = \to c \cdot \to a = 0$
Angle between $\to a$ and $\to a + \to b + \to c$ is
$cos θ = \frac{\to a (\to a + \to b + \to c)}{| \to a | ∣ ∣ \to a + \to b + \to c ∣ ∣}$ .......(1)
Now $| \to a | = ∣ ∣ \to b ∣ ∣ = | \to c | = a$
$∣ ∣ \to a + \to b + \to c ∣ ∣ = {\to a}^{2} + {\to b}^{2} + {\to c}^{2} + 2 (\to a . \to b) + 2 (\to b . \to c) + 2 (\to c . \to a) = {\to a}^{2} + {\to b}^{2} + {\to c}^{2} + 0 + 0 + 0$
$\Rightarrow {∣ ∣ \to a + \to b + \to c ∣ ∣}^{2} = 3 a^{2} \Rightarrow ∣ ∣ \to a + \to b + \to c ∣ ∣ = \sqrt{3} a$
Substitute this value in (1) to get, $θ = {cos}^{- 1} (\frac{1}{\sqrt{3}})$

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