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Dot Product

Let p, q, r...

Question

Let $p, q, r$ be three mutually perpendicular vectors of the same magnitude. If a vector $x$ satisfies the equation $p x \times ((x - q) \times p) + q \times ((x - r) \times q) + r \times ((x - p) \times r) = 0$ then $x$ is given by

A

\frac{1}{2} (p + q - 2 r)

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B

\frac{1}{2} (p + q + r)

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C

\frac{1}{3} (p + q - 2 r)

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D

\frac{1}{3} (2 p + q - r)

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Solution

The correct option is C $\frac{1}{2} (p + q + r)$
Let $∣ ∣ \to p ∣ ∣ = ∣ ∣ \to q ∣ ∣ = ∣ ∣ \to r ∣ ∣ = K$ . Let $^p,^q,^r$ be unit vectors along $\to p, \to q, \to r$ respectively.
Clearly $^p,^q,^r$ are mutually perpendicular vectors, so any vector $\to x$ can be written as $a_{1}^p + a_{2}^q + a_{3}^r$ .
$\to p \times ((\to x - \to q) \times \to p) = (\to p . \to p) (\to x - \to q) - (\to p . (\to x - \to q)) \to p = K^{2} (\to x - \to q) - (\to p . \to x) \to p (\to p . \to q = 0) = K^{2} (\to x - \to q) - K^p . (a_{1}^p + a_{2}^q + a_{3}^r) K^p = (\to x - \to q - a_{1}^p)$
Similarly $\to q \times ((\to x - \to r) \times \to q) = K^{2} (\to x - \to r - a_{2}^q)$ and $\to r \times ((\to x - \to p) \times \to r) = K^{2} (\to x - \to p - a_{3}^r)$
According to the given condition
$K^{2} (\to x - \to q - a_{1}^p + \to x - \to r - a_{2}^q + \to x - \to r - a_{3}^r) = 0 \Rightarrow K^{2} (3 \to x - (\to p + \to q + \to r) - (a_{1}^p + a_{2}^q + a_{3}^r)) = 0 \Rightarrow K^{2} (2 \to x - (\to p + \to q + \to r)) = 0 \Rightarrow \to x = \frac{1}{2} (\to p + \to q + \to r) (∵ K \neq 0)$

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