Let a-b-c be non zero vectors such that a-b -c-13bca- If - is the acute angle between the vectors b and c then sin- equals

The correct option is D 2√(2)3 (a×b )×c=13| b| |c| | a|..... ( As given ) ⇒(a·c)b (b·c)a =13|b| |c| |a| #160; ⇒ b· #160;c=13|b| |c ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

Let a,b,c b...

Question

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

\frac{2}{3}

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

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\frac{1}{3}

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

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Solution

The correct option is D $\frac{2 \sqrt{2}}{3}$
$(\to a \times \to b) \times \to c = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ ∣ ∣ \to a ∣ ∣ . . . . . ($ As given $)$

$\Rightarrow (\to a \cdot \to c) \to b - (\to b \cdot \to c) \to a = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ ∣ ∣ \to a ∣ ∣$

$\Rightarrow - \to b \cdot \to c = \frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣$

$\Rightarrow cos θ = - \frac{1}{3}$

$sin θ = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

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Let →a,→b,→c be non zero vectors such that (→a×→b)×→c=13∣∣→b∣∣∣∣→c∣∣∣∣→a∣∣. If θ is the acute angle between the vectors →b and →c then sinθ equals