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Orthogonal System of Vectors

Let p , q , r...

Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is B $\frac{1}{2} (\to p + \to q + \to r)$
Since, $\to p, \to q, \to r$ are mutually perpendicular vectors of same magnitude, so let us consider
$\begin{matrix} | \to p | = | \to q | = | \to r | = λ a n d \to p . \to q = \to q . \to r = \to r . \to p = 0 \end{matrix}}$ ...(i)
Given, $\to p \times {(\to x - \to q) \times \to p} + \to q \times {(\to x - \to r) \times \to q} + \to r \times {(\to x - \to p) \times \to r} = \to 0$
$\Rightarrow (\to p . \to p) (\to x - \to q) - {\to p . (\to x - \to q)} \to p + (\to q . \to q) (\to x - \to r) - {\to q . (\to x - \to r)} \to q + (\to r . \to r) (\to x - \to p) - {\to r . (\to x - \to p)} \to r = 0$
$\Rightarrow \to x {(\to p . \to p) + (\to q . \to q) + (\to r . \to r)} - (\to p . \to p) \to q - (\to q . \to q) r - (\to r . \to r) \to p = (\to x . \to p) \to p + (\to x . \to q) \to q + (\to x . \to r) \to r$
$\Rightarrow 3 \to x | λ |^{2} - (\to p + \to q + \to r) | λ |^{2} = (\to x . \to p) \to p + (\to x . \to q) \to q + (\to x . \to r) \to r$ ...(ii)
Taking dot of Eq. (ii) with $\to p$ we get
$3 (\to x . \to p) | λ |^{2} - | λ |^{4} = (\to x . \to p) | λ |^{2} \Rightarrow= \to x . \to p = \frac{1}{2} | λ |^{2}$
Similarly, taking dot of Eq. (ii) with $\to q$ and $\to r$ respectively, we get
$\to x . \to q = \frac{| λ |^{2}}{2} = \to x . \to r$
$∴$ Eq. (ii) becomes
$3 \to x | λ |^{2} - (\to p + \to q + \to r) | λ |^{2} = \frac{| λ |^{2}}{2} (\to p + \to q + \to r)$
$\Rightarrow 3 \to x = \frac{1}{2} (\to p + \to q + \to r) + (\to p + \to q + \to r)$
$\Rightarrow \to x = \frac{1}{2} (\to p + \to q + \to r)$

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