If the resultant of any two vectors A→ and B→ is numerically equal to the either vector, the angle between the vectors can be:

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If the resultant of any two vectors $\vec{A} and \vec{B}$ is numerically equal to the either vector, the angle between the vectors can be:

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\cos^{- 1} (- \frac{B}{2A}) and \cos^{- 1} (- \frac{A}{2B})

\sin^{- 1} (\frac{A}{B}) and \cos^{- 1} (- \frac{B}{2A})

\sin^{- 1} (- \frac{A}{B}) and \cos^{- 1} (- \frac{2A}{B})

none of these

answer is A.

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Detailed Solution

l e t \bar{A} + \bar{B} = \bar{R}

$R^{2} = A^{2} + B^{2} + 2 AB \cos θ$

Where $θ \to$ angle between $\bar{A} and \bar{B}$

$if R = A then θ = \cos^{- 1} (\frac{- B}{2 A})$

$if R = B then θ = \cos^{- 1} (\frac{- A}{2 B})$

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