Let the angle between two non-zero vectors A and B be 1200 and resultant be C- then

Given, two non zero vectors A and B, angle between vectors (θ ) = nbsp;120^^0 As we know that the resultant of two vectors C = nbsp;A + B or nbsp;| C | = ...

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Standard XII

Physics

Vectors

Let the angle...

Question

Let the angle between two non-zero vectors $\to A$ and $\to B$ be $120^{^{0}}$ and resultant be $\to C$ , then

∣ ∣ ∣ \to C ∣ ∣ ∣

must be equal to

∣ ∣ ∣ \to A - \to B ∣ ∣ ∣

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∣ ∣ ∣ \to C ∣ ∣ ∣

must be less than

∣ ∣ ∣ \to A - \to B ∣ ∣ ∣

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∣ ∣ ∣ \to C ∣ ∣ ∣

must be greater than

∣ ∣ ∣ \to A - \to B ∣ ∣ ∣

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∣ ∣ ∣ \to C ∣ ∣ ∣

may be equal to

∣ ∣ ∣ \to A - \to B ∣ ∣ ∣

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Solution

The correct option is B $∣ ∣ ∣ \to C ∣ ∣ ∣$ must be less than $∣ ∣ ∣ \to A - \to B ∣ ∣ ∣$
Given,
two non-zero vectors
$\to A$ and $\to B$ , angle between vectors $(θ)$ = $120^{^{0}}$
As we know that the resultant of two vectors
$\to C$ = $\to A + \to B$
or $∣ ∣ ∣ \to C ∣ ∣ ∣ = ∣ ∣ ∣ \to A + \to B ∣ ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
or $∣ ∣ ∣ \to C ∣ ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos (120^{o})}$
$= \sqrt{A^{2} + B^{2} - A B}$
$∣ ∣ ∣ \to C ∣ ∣ ∣ < ∣ ∣ ∣ \to A - \to B ∣ ∣ ∣$

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Let the angle between two non-zero vectors →A and →B be 1200 and resultant be →C, then

Let the angle between two non-zero vectors $\to A$ and $\to B$ be $120^{^{0}}$ and resultant be $\to C$ , then