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

We have 13|b | |c | a=(a×b)×c We know that (a×b)×c=(a·c)b (b·c)a ∴ nbsp;13 |b||c|a=(a·c)b (b·c)a on comparing both sides, we get ⇒b·c= 13|b||c| ∴cosθ= 13⇒sinθ ...

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Byju's Answer

Standard XII

Mathematics

Test for Collinearity of Vectors

Let a,b and c...

Question

Let $\to a, \to b$ and $\to c$ be non-zero vectors such that no two 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 the vectors $\to b$ and $\to c$ , then the value of sin $θ$ is

\frac{- 1}{3}

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\frac{- 2 \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 $\frac{2 \sqrt{2}}{3}$
We have $\frac{1}{3} | \to b | | \to c | \to a = (\to a \times \to b) \times \to c$
We know that $(\to a \times \to b) \times \to c = (\to a \cdot \to c) \to b - (\to b \cdot \to c) \to a$ $∴ \frac{1}{3} | \to b | | \to c | \to a = (\to a \cdot \to c) \to b - (\to b \cdot \to c) \to a$
on comparing both sides, we get
$\Rightarrow \to b \cdot \to c = - \frac{1}{3} | \to b | | \to c |$
$∴ cos θ = - \frac{1}{3} \Rightarrow sin θ = \frac{2 \sqrt{2}}{3}$

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Test for Collinearity of Vectors

Standard XII Mathematics

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

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