Let the angle between two nonzero vectors A→ and B→ be 120° and its resultant be C→.

R=A2+B2+2ABcos120oC=A2+B2-ABA→-B→=A2+B2-2ABcos120o⇒A→-B→=A2+B2+ABC<A→-B→

Let the angle between two nonzero vectors A→ and B→ be 120° and its resultant be C→.

R=A2+B2+2ABcos120oC=A2+B2-ABA→-B→=A2+B2-2ABcos120o⇒A→-B→=A2+B2+ABC<A→-B→

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Let the angle between two nonzero vectors $\vec{A} and \vec{B}$ be 120° and its resultant be $\vec{C}$ .

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C must be equal to $| \vec{A} - \vec{B} |$

C must be less than $| \vec{A} - \vec{B} |$

C must be greater than $| \vec{A} - \vec{B} |$

C may be equal to $| \vec{A} - \vec{B} |$

answer is B.

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Detailed Solution

$R = \sqrt{A^{2} + B^{2} + 2 ABcos 120^{o}}$

$C = \sqrt{A^{2} + B^{2} - AB}$

$|\vec{A} - \vec{B}| = \sqrt{A^{2} + B^{2} - 2 ABcos 120^{o}}$

$\Rightarrow |\vec{A} - \vec{B}| = \sqrt{A^{2} + B^{2} + AB}$

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

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Let the angle between two nonzero vectors A→ and B→ be 120° and its resultant be C→.

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Let the angle between two nonzero vectors $\vec{A} and \vec{B}$ be 120° and its resultant be $\vec{C}$ .