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It is found that $| A + B | = | A |$ . This necessarily implies.

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B = 0

A, B are perpendicular

A, B are anti parallel

A .B \leq 0

answer is D.

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Detailed Solution

| \vec{A} + \vec{B} | = | \vec{A} | S .O .B .S

$A^{2} + B^{2} + 2 AB \cos θ = A^{2}$

$B^{2} + 2 AB \cos θ = 0$

$2 AB \cos θ = - B^{2}$

$AB \cos θ = \frac{- B^{2}}{2}$

$\vec{A} . \vec{B} = \frac{- B^{2}}{2} \Rightarrow \vec{A} . \vec{B} \leq 0$

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