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 C 2√(2)3 Given: (a⃗×b⃗) ×c⃗ = 13 | b⃗ | | c⃗ |a⃗ (a⃗×b⃗) ×c⃗ = a⃗(b⃗.c⃗)+b⃗(a⃗.c⃗)= a⃗ | b⃗ | | c⃗ |cosθ+b⃗ | a⃗ | | c⃗ | ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

Let a⃗, b⃗a...

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{2}}{3}

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Solution

The correct option is C $\frac{2 \sqrt{2}}{3}$
Given:
$(\to a \times \to b) \times \to c = \frac{1}{3} ∣ ∣ \to b ∣ ∣ | \to c | \to a$

$(\to a \times \to b) \times \to c = - \to a (\to b . \to c) + \to b (\to a . \to c) = - \to a ∣ ∣ \to b ∣ ∣ | \to c | cos θ + \to b | \to a | | \to c | cos α$ .......... $(α$ is the angle between $a$ and $c)$

$- \to a ∣ ∣ \to b ∣ ∣ | \to c | cos θ + \to b | \to a | | \to c | cos α = \frac{1}{3} ∣ ∣ \to b ∣ ∣ | \to c | \to a$

Comparing both the sides.
We get, $cos α = 0$
hence,

$- \to a ∣ ∣ \to b ∣ ∣ | \to c | cos θ = \frac{1}{3} ∣ ∣ \to b ∣ ∣ | \to c | \to a$

$⟹ cos θ = - \frac{1}{3}$

We have,
${cos}^{2} θ + {sin}^{2} θ = 1$
$sin θ = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - \frac{1}{9}} = \frac{\sqrt{8}}{3} = \frac{2 \sqrt{2}}{3}$

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Let →a,→band →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 sin θ is :

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 :