Let a-b-c be nonzero vectors such that a-b-c-13-b- -c-a- If - is the angle between b and c- then sin-

The correct option is D 2√(2)3 13|b| |c|a = (a×b)×c =(a.c)b (b.c)a ⇒a.c=0, b.c= 13|b| |c| cosθ= 13 sinθ=√(1 cos^2θ) nbsp; nb ...

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Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

Let a,b,c b...

Question

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

\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 $\frac{2 \sqrt{2}}{3}$
$\frac{1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣ \to a = (\to a \times \to b) \times \to c$
$= (\to a . \to c) \to b - (\to b . \to c) \to a$
$\Rightarrow \to a . \to c = 0, \to b . \to c = \frac{- 1}{3} ∣ ∣ ∣ \to b ∣ ∣ ∣ ∣ ∣ \to c ∣ ∣$
$cos θ = \frac{- 1}{3}$
$sin θ = \sqrt{1 - {cos}^{2} θ}$
$= \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

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Similar questions

Q. Let

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k,

and

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero vectors such that

c

is unit vector perpendicular to both vectors

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b

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\frac{π}{6}

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{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

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Q. Let

a = a_{1} i + a_{2} j + a_{3} k, b = b_{1} i + b_{2} j + b_{3} k

and

c = c_{1} i + c_{2} j + c_{3} k

be three non-zero such that

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\frac{π}{6},

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{∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣}^{2}

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Q. Let

\to a = a_{1}^i + a_{2}^j + a_{3}^k, \to b = b_{1}^i + b_{2}^j + b_{3}^k

and

\to c = c_{1}^i + c_{2}^j + c_{3}^k

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{⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦}^{2} =

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

The correct option is D 2√2313∣∣∣→b∣∣∣∣∣→c∣∣→a=(→a×→b)×→c=(→a.→c)→b−(→b.→c)→a⇒→a.→c=0,→b.→c=−13∣∣∣→b∣∣∣∣∣→c∣∣cosθ=−13sinθ=√1−cos2θ =√1−19=√89=2√23

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