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Sin(A+B)Sin(A-B)

If a⃗,b⃗ an...

Question

If $\to a, \to b$ and $\to c$ are unit vectors such that: $\to a + 2 \to b + 2 \to c = \to 0$ , then $| \to a \times \to c |$ is equal to :

A

\frac{\sqrt{15}}{4}

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B

\frac{1}{4}

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C

\frac{15}{16}

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D

\frac{\sqrt{15}}{16}

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Solution

The correct option is A $\frac{\sqrt{15}}{4}$
$\begin{matrix} \to a + 2 \to b + 2 \to c = \to 0 a n d | \to a | = ∣ ∣ \to b ∣ ∣ = | \to c | = 1 \to a + 2 \to c = - 2 \to b {| \to a + 2 \to c |}^{2} = 2 {| \to a + 2 \to c |}^{2} = 4 (\to a + 2 \to c) . (\to a + 2 \to c) = 4 1 + 4 + 4 \to a . \to c = 4 \to a . \to c = \frac{- 1}{4} cos θ = \frac{- 1}{4} \Rightarrow sin θ = \frac{\sqrt{15}}{4} a n d, | \to a \times \to c | = | \to a | | \to c | sin θ = \frac{\sqrt{15}}{4} h e n c e, t h e o p t i o n A i s c o r r e c t . \end{matrix}$

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