Let a- b- and c- be non-zero vectots such that a- b- c- - 13 - b- - - c- -a- - If - is the acute angle between the b- and c- - then sin - equals

The correct option is D 2√(2)3 Given, #160; (a⃗×b⃗) ×c⃗=1/3|b⃗||c⃗| a⃗ ∵(a⃗× b) ×c⃗ can be written as ⃗c⃗ (a⃗×b⃗) ∴ C⃗×(a⃗×b⃗)=1/3| ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Lines

Let a⃗, b...

Question

Let $\to a$ , $\to b$ and $\to c$ be non-zero vectots such that ( $\to a$ $\to b$ ) $\to c$ = $\frac{1}{3}$ $∣ ∣ \to b ∣ ∣ | \to c | \to a$ . If $θ$ is the acute angle between the $\to b$ and $\to c$ , then sin $θ$ equals

\frac{2}{3}

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

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\frac{1}{3}

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

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Solution

The correct option is D $\frac{2 \sqrt{2}}{3}$
Given,
$(\to a \times \to b) \times \to c = \frac{1}{3} | \to b | | \to c | \to a$
$∵ (\to a \times b) \times \to c$ can be written as $\to - c (\to a \times \to b)$
$∴ - \to C \times (\to a \times \to b) = \frac{1}{3} | \to b | | \to c | \to a$
$- (\to c \cdot \to b) a + (\to c \cdot \to a) \to b = \frac{1}{3} | \to b | | \to c | \to a$
$(\to c \cdot \to a) \to b = [\frac{1}{3} | \to b | \to c ∣ + \to c \cdot \to b] \to a$

$∴$ given, $\to a$ and $¯ b$ are not collenear
$c \cdot b + \frac{1}{3} | \to b | \to c ∣= 0$ and $\to c \cdot \to a = 0$
$| c | | b | cos θ + \frac{1}{3} \to b / | \to c | = 0$
$| \to b | \to c ∣ (cos θ + \frac{1}{3}) = 0$
$\Rightarrow cos θ = \frac{- 1}{3}$
Then, $sin θ = \frac{\sqrt{8}}{3} = \frac{2 \sqrt{2}}{3}$

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Let →a, →b and →c be non-zero vectots such that (→a →b) →c = 13∣∣→b∣∣|→c|→a . If θ is the acute angle between the →b and →c , then sinθ equals

Let $\to a$ , $\to b$ and $\to c$ be non-zero vectots such that ( $\to a$ $\to b$ ) $\to c$ = $\frac{1}{3}$ $∣ ∣ \to b ∣ ∣ | \to c | \to a$ . If $θ$ is the acute angle between the $\to b$ and $\to c$ , then sin $θ$ equals