First slide

Dot product or scalar product of vectors

Question

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

Moderate

A

2
B

-1
C

1
D

$\sqrt{6} / 3$

Solution

$\begin{array}{l} | \vec{a} + \vec{b} + \vec{c} |^{2} = 6 \\ or | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} \\ + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 \\ \Rightarrow | \vec{a} | = | \vec{b} | = | \vec{c} | and \\ \vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cdot \cos \frac{π}{3} \\ i.e., \vec{a} \cdot \vec{b} = \frac{1}{2} | \vec{a} |^{2} \\ ∴ 3 | \vec{a} |^{2} + 3 | \vec{a} |^{2} = 6 \\ \Rightarrow | \vec{a} |^{2} or | \vec{a} | = 1 \end{array}$

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