If (A→+B→) is perpendicular to B→ and (A→+2B→) is perpendicular to A→, then

(A→+B→).B→=0⇒A→.B→+B2=0→(1) (A→+2B→).A→=0⇒A2+2A→.B→=0→(2) From (1) and (2), A=2B

If (A→+B→) is perpendicular to B→ and (A→+2B→) is perpendicular to A→, then

(A→+B→).B→=0⇒A→.B→+B2=0→(1) (A→+2B→).A→=0⇒A2+2A→.B→=0→(2) From (1) and (2), A=2B

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If $(\vec{A} + \vec{B})$ is perpendicular to $\vec{B}$ and $(\vec{A} + 2 \vec{B})$ is perpendicular to $\vec{A},$ then

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$2 A = B$

$A = \sqrt{2} B$

$A = 2 B$

$A = B$

answer is A.

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Detailed Solution

$(\vec{A} + \vec{B}) . \vec{B} = 0 \Rightarrow \vec{A} . \vec{B} + B^{2} = 0 \to (1)$
$(\vec{A} + 2 \vec{B}) . \vec{A} = 0 \Rightarrow A^{2} + 2 \vec{A} . \vec{B} = 0 \to (2)$
From (1) and (2), $A = \sqrt{2} B$

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If (A→+B→) is perpendicular to B→ and (A→+2B→) is perpendicular to A→, then

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If $(\vec{A} + \vec{B})$ is perpendicular to $\vec{B}$ and $(\vec{A} + 2 \vec{B})$ is perpendicular to $\vec{A},$ then