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Applications of Dot Product

a⃗, b⃗ and ...

Question

$\to a, \to b$ and $\to c$ are three vectors of equal magnitude. 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{\sqrt{6}}{3}

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Solution

The correct option is C $1$
We have
$| \to a + \to b + \to c |^{2} = 6$
or $| \to a |^{2} + | \to b |^{2} + | \to c |^{2}$
$+ 2 (\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a) = 6$
$\Rightarrow | \to a | = | \to b | = | \to c |$ and $\to a \cdot \to b$
$= | \to a | | \to b | \cdot cos \frac{π}{3}$
i.e., $\to a \cdot \to b = \frac{1}{2} | \to a |^{2}$
$∴ 3 | \to a |^{2} + 3 | \to a |^{2} = 6$
$\Rightarrow | \to a |^{2}$ or $| \to a | = 1$

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