Let the angle between two nonzero vectors A- and B- be 120o and resultant be C-

The correct option is C C⃗ must be less than | A⃗ B⃗| As we know, C=√(A^2+B^2+2ABcosθ) cos(120^0)= 0.5 so, R=√(A^2+B^2 AB ) And | ...

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Byju's Answer

Standard XII

Physics

Vector Component

Let the angle...

Question

Let the angle between two nonzero vectors $\to A$ and $\to B$ be $120^{o}$ and resultant be $\to C$

\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 C $\to C$ must be less than $| \to A - \to B |$
As we know,
$C = \sqrt{A^{2} + B^{2} + 2 A B c o s θ}$
$c o s (120^{0}) = - 0.5$
so,
$R = \sqrt{A^{2} + B^{2} - A B}$

And $| \to A - \to B | = \sqrt{A^{2} + B^{2} - 2 A B c o s θ}$
so,
$| \to A - \to B | = \sqrt{A^{2} + B^{2} + A B}$

From the above result
option B is correct.

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Let the angle between two nonzero vectors →A and →B be 120o and resultant be →C

The correct option is C →C must be less than |→A−→B|As we know,C=√A2+B2+2ABcosθcos(1200)=−0.5so,R=√A2+B2−ABAnd |→A−→B|=√A2+B2−2ABcosθso,|→A−→B|=√A2+B2+ABFrom the above result option B is correct.

Let the angle between two nonzero vectors $\to A$ and $\to B$ be $120^{o}$ and resultant be $\to C$