First slide

Vectors

Question

Two vectors $\vec{a} and \vec{b}$ of different magnitude are found to have a resultant of magnitude R. On reversing one of them, the resultant is again found to have the same magnitude R. Then the angle between the vectors $\vec{a} and \vec{b}$ must be

Easy

A

$90^{0}$
B

$120^{0}$
C

$60^{0}$
D

$45^{0}$

Solution

Given $|\vec{a} + \vec{b}| = R$ , on reversing the second vector $|\vec{a} - \vec{b}| = R$ . Thus $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$ . If $θ$ is angle between $\vec{a} and \vec{b}$ , then $\sqrt{a^{2} + b^{2} + 2 abcosθ} = \sqrt{a^{2} + b^{2} - 2 abcosθ} .$ Solve for $θ$

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