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Angle between Two Lines

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ are three vectors of equal magnitude and the angle between each pair of vectors is $\frac{π}{3}$ such that $∣ ∣ \to a + \to b + \to c ∣ ∣ = \sqrt{6}$ then $| \to a |$ is equal to

A

2

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B

- 1

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C

1

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D

\frac{1}{3} \sqrt{6}

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Solution

The correct option is C $1$
Let $\to x = \to b + \to c$
$\Rightarrow x^{2} = b^{2} + c^{2} + 2 b c (cos 60^{o}) = b^{2} + c^{2} + b c$

$| \to a + \to b + \to c | = | \to a + \to x | = \sqrt{a^{2} + x^{2} + 2 \to a \cdot \to x}$

$| \to a + \to b + \to c | = \sqrt{a^{2} + b^{2} + c^{2} + b c + a b + a c} = \sqrt{6 a^{2}}$
$(∵$ the magnitudes of all the vectors is equal $)$

As per the given information, $\sqrt{6} = \sqrt{6 a^{2}}$

$\Rightarrow | \to a | = 1$

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