Let a- b- and c- be non-zero vectors such that a-b-c-1-3-b-c-a- If - is the acute angle between the vectors b and c- then sin- equals-

The correct option is D 2√(2)/3 Since #160; 1 / 3 | b #160;| | c #160;| a =( a × b #160;) × c We know that ( a × b #160;) × c ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Let a⃗, b⃗ ...

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 $¯ ¯ ¯ b$ and $¯ ¯ ¯ c$ , then $sin θ$ equals:

\frac{1}{3}

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

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\frac{2}{3}

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

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Solution

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

We know that $(\to a \times \to b) \times \to c = (\to a . \to c) \to b - (\to b . \to c) \to a$

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

On comparing we get

$\frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ = - \to b . \to c$ and $\to a . \to c = 0$

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

$\Rightarrow {cos}^{2} θ = \frac{1}{3} \Rightarrow 1 - {sin}^{2} θ = \frac{1}{9}$

$\Rightarrow {sin}^{2} θ = 1 - \frac{1}{9} = \frac{8}{9}$

$\Rightarrow sin θ = \frac{2 \sqrt{2}}{3}$

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Let →a, →b and →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:

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