Let $\to a, \to b, \to c$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $θ$ , with the vector $\to a + \to b + \to c .$ Then $36 {cos}^{2} 2 θ$ is equal to

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Solution

Let $∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = t$
Since $\to a, \to b, \to c$ are mutually perpendicular.
So, $\to a \cdot \to b = \to b \cdot \to c = \to c \cdot \to a = 0$

${∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣}^{2} = {∣ ∣ \to a ∣ ∣}^{2} + {∣ ∣ ∣ \to b ∣ ∣ ∣}^{2} + {∣ ∣ \to c ∣ ∣}^{2} + 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a)$
$\Rightarrow {∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣}^{2} = 3 t^{2}$
$\Rightarrow ∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣ = \sqrt{3} t$
Now,
$cos θ = \frac{\to a \cdot (\to a + \to b + \to c)}{∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to a + \to b + \to c ∣ ∣ ∣} = \frac{{∣ ∣ \to a ∣ ∣}^{2}}{∣ ∣ \to a ∣ ∣ (\sqrt{3} ∣ ∣ \to a ∣ ∣)} = \frac{t^{2}}{t^{2} \sqrt{3}} = \frac{1}{\sqrt{3}}$
So, $36 {cos}^{2} 2 θ = 36 (2 {cos}^{2} θ - 1)^{2}$
$= 36 {(\frac{2}{3} - 1)}^{2} = 36 \times \frac{1}{9} = 4$

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Let $\to a, \to b, \to c$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $θ$ , with the vector $\to a + \to b + \to c .$ Then $36 {cos}^{2} 2 θ$ is equal to