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

JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

Let

\overline{a}, \overline{b}

and

\overline{c}

be non-zero vectors such that

(\overline{a} \times \overline{b}) \times \overline{c} = \frac{1}{3} ∣ \overline{b} ∣∣ \overline{c} ∣ \overline{a}

. If

θ

is the acute angle between the vectors

\overline{b}

and

\overline{c}

, then

sin θ

equals

Answer: (D)

The vector triple product expands as

(a \times b) \times c = (a \cdot c) b - (b \cdot c) a

. Matching the coefficient of

a

against the right-hand side

\frac{1}{3} ∣ b ∣∣ c ∣ a

gives

- (b \cdot c) = \frac{1}{3} ∣ b ∣∣ c ∣

, so

∣ b ∣∣ c ∣ cos θ = - \frac{1}{3} ∣ b ∣∣ c ∣

and

cos θ = - \frac{1}{3}

. For the acute angle between the vectors

∣ cos θ ∣ = \frac{1}{3}

, hence

sin θ = 1 - \frac{1}{9} = \frac{2 2}{3}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option