Let a- b- and c- be three non-zero vectors such that no two of them are collinear and a-b-c-13-b-c-a- If - is the angle between vectors b- and c- then a value of sin- is

The correct option is B 2√(2)3 (a⃗× #160;b⃗)× #160;c⃗ = (a⃗ . c⃗) b⃗ (b⃗ . #160;c⃗) a⃗ = 13|b⃗||c⃗|a⃗ By comparing, #160; b⃗ . c⃗= 13| ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

Let a⃗, b⃗ ...

Question

Let $\to a, \to b$ and $\to c$ be three non-zero vectors such that no two of them are collinear and $(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a$ . If $θ$ is the angle between vectors $\to b$ and $\to c$ , then a value of $sin θ$ is

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

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

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\frac{2}{3}

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

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Solution

The correct option is B $\frac{2 \sqrt{2}}{3}$
$(\to a \times \to b) \times \to c = (\to a . \to c) \to b - (\to b . \to c) \to a = \frac{1}{3} | \to b | | \to c | \to a$
By comparing, $\to b . \to c = - \frac{1}{3} | \to b | | \to c |$
$\Rightarrow cos θ = \frac{- 1}{3}$
$\Rightarrow sin θ = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

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Let →a,→b and →c be three non-zero vectors such that no two of them are collinear and (→a×→b)×→c=13|→b||→c|→a. If θ is the angle between vectors →b and →c, then a value of sinθ is

The correct option is B 2√23(→a×→b)×→c=(→a.→c)→b−(→b.→c)→a=13|→b||→c|→aBy comparing, →b.→c=−13|→b||→c|⇒cosθ=−13⇒sinθ=√1−cos2θ=√1−19=√89=2√23

Let $\to a, \to b$ and $\to c$ be three non-zero vectors such that no two of them are collinear and $(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a$ . If $θ$ is the angle between vectors $\to b$ and $\to c$ , then a value of $sin θ$ is