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Scalar Triple Product

Let a, b, c b...

Question

Let $\to a, \to b, \to c$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\to r$ satisfies
$\to a \times {(\to r - \to b) \times \to a} + \to b \times {(\to r - \to c) \times \to b} + \to c \times {(\to r - \to a) \times \to c} = \to 0$
then $\to r$ is equal to

A

\frac{1}{3} (\to a + \to b + \to c)

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B

\frac{1}{2} (\to a + \to b + 2 \to c)

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C

\frac{1}{2} (\to a + \to b + \to c)

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D

\frac{1}{3} (2 \to a + \to b - \to c)

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Solution

The correct option is C $\frac{1}{2} (\to a + \to b + \to c)$
$∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣ = d$ (say)
$\to a \cdot \to b = \to b \cdot \to c = \to c \cdot \to a = 0$
$\to a \times {(\to r - \to b) \times \to a} + \to b \times {(\to r - \to c) \times \to b} + \to c \times {(\to r - \to a) \times \to c} = \to 0$
$\sum \to a \times {(\to r - \to b) \times \to a} = \to 0 \Rightarrow \sum (a \cdot a) (\to r - \to b) - (\to a \cdot (\to r - \to b)) \to a = \to 0 \Rightarrow \sum d^{2} (\to r - \to b) - (\to a \cdot \to r) \to a = \to 0 [∵ \to a \cdot \to b = \to 0]$
Given that $\to a, \to b, \to c$ are three vectors mutually perpendicular to each other.
Therefore, let $\to r = r_{1} \to a + r_{2} \to b + r_{3} \to c$
$∴ \sum d^{2} (\to r - \to b) - (r_{1} d^{2}) \to a = \to 0 [∵ \to a \cdot \to r = r_{1} {∣ ∣ \to a ∣ ∣}^{2}] \Rightarrow 3 d^{2} \to r - d^{2} (\to a + \to b + \to c) - d^{2} {r_{1} \to a + r_{2} \to b + r_{3} \to c} = \to 0 \Rightarrow 3 d^{2} \to r - d^{2} (\to a + \to b + \to c) - d^{2} \to r = \to 0 \Rightarrow 2 \to r - (\to a + \to b + \to c) = \to 0 \Rightarrow \to r = \frac{1}{2} (\to a + \to b + \to c)$

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