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Cross Product of Two Vectors

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Question

Let $\to a, \to b$ and $\to c$ be non-zero vectors such that $(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a$ . If $θ$ is the acute angle between the vectors $\to b$ and $\to c$ , then $sin θ$ equals?

A

\frac{1}{3}

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B

\frac{\sqrt{2}}{3}

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C

\frac{2}{3}

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D

\frac{2 \sqrt{2}}{3}

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Solution

The correct option is D $\frac{2 \sqrt{2}}{3}$
$\begin{matrix} (\to a \times \to b) \times \to c = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a - \to c (\to a \times \to b) = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a - [(\to c \times \to b) \to a - (\to c \times \to a) \to b] = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a (\to c \times \to a) \to b - (\to c \times \to b) \to a = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a - (\to c \times \to b) \to a = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a - ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a cos θ = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a cos θ = - \frac{1}{3} sin θ = \frac{2 \sqrt{2}}{3} \end{matrix}$
Option $D$ is correct.

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