Let the angle between two nonzero vectors A and B be 120^° and re… - Vidyadip

MCQ

Let the angle between two nonzero vectors $\overrightarrow A $ and $\overrightarrow B $ be $120^°$ and resultant be $\overrightarrow C $

A
$\overrightarrow C $ must be equal to $|\overrightarrow A - \overrightarrow B |$
B
$\overrightarrow C $ must be greater than $|\overrightarrow A - \overrightarrow B |$
✓
$\overrightarrow C $ must be less than $|\overrightarrow A - \overrightarrow B |$
D
$\overrightarrow C $ may be equal to $|\overrightarrow A - \overrightarrow B |$

✓

Answer

Correct option: C.

$\overrightarrow C $ must be less than $|\overrightarrow A - \overrightarrow B |$

c
(c) If $\overrightarrow{\mathrm{C}}$ is resultant of $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$, then

$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos 120^{\circ}}$

$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-\mathrm{AB}} \quad\left[\text { Ascos } 120^{\circ}=-\frac{1}{2}\right]$

similarly, $|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos 120^{\circ}}$

$=\sqrt{A^{2}+B^{2}+A B}$

$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|>\mathrm{C}$

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