First slide

Vectors

Question

The angle between two vectors $\vec{A} and \vec{B}$ is 120⁰, then its resultant C will be

Moderate

A

$C = |\vec{A} - \vec{B}|$
B

$C > |\vec{A} - \vec{B}|$
C

$C = |- \vec{A} + \vec{B}|$
D

$C < |\vec{A} - \vec{B}|$

Solution

$C = \sqrt{A^{2} + B^{2} + 2 A B \cos 120^{0}} = \sqrt{A^{2} + B^{2} - A B} a n d |\vec{A} - \vec{B}| = \sqrt{A^{2} + B^{2} - 2 A B \cos 120^{0}} = \sqrt{A^{2} + B^{2} + A B} C o m p a r i n g t h e m, w e c a n s a y C < | \vec{A} - \vec{B} |$

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