If the angle between two vectors A and B is $120 °$ , then its resultant C will be

$C = |A - B|$

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$C < |A - B|$

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$C > |A - B|$

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$C = |A + B|$

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Solution

The correct option is B
$C < |A - B|$

Step 1: Given data:
Two vectors $\vec{A}$ and $\vec{B}$ .
Angle between A and B: $θ = 120 °$
Step 2: Formula Used
Resultant of 2 vectors is $C = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
Difference of 2 vectors is $|A - B| = \sqrt{A^{2} + B^{2} - 2 A B \cos θ}$
Step 3: Solution
Calculate the resultant C. Here, $c o s θ = \cos 120 = - \frac{1}{2}$
$\begin{array}{rcl} C & = & \sqrt{A^{2} + B^{2} + 2 A B \cos θ} \\ = & \sqrt{A^{2} + B^{2} + 2 A B \cos 120} \\ = & \sqrt{A^{2} + B^{2} + 2 A B (- \frac{1}{2})} \\ = & \sqrt{A^{2} + B^{2} - A B} \end{array}$
Calculate the difference. Here, $c o s θ = \cos 120 = - \frac{1}{2}$
$\begin{array}{rcl} |A - B| & = & \sqrt{A^{2} + B^{2} - 2 A B \cos θ} \\ = & \sqrt{A^{2} + B^{2} - 2 A B \cos 120} \\ = & \sqrt{A^{2} + B^{2} - 2 A B (- \frac{1}{2})} \\ = & \sqrt{A^{2} + B^{2} + A B} \end{array}$
Since $\begin{array}{rcl} \sqrt{A^{2} + B^{2} + A B} \end{array} > \begin{array}{rcl} \sqrt{A^{2} + B^{2} - A B} \end{array}$ , therefore $|A - B| > C$ .
Hence, resultant C will be $C < |A - B|$

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