If A⇀+B⇀=R⇀ and 2A⇀+B⇀ is perpendicular to B⇀ then

A→+B→=R→ A→+B→=R given 2A→+B→ is perpendicular to B→ (2A→+B→)⋅B→=0⇒[A→+(A→+B→)]⋅B→=0(A→+R→)⋅B→=0⇒A→=−R→∴A=R [∵B≠0 and R→ cannot be perpendicular to B→]Alternatively 2A⇀+B →is perpendicular to B cosθ=-B2AR2=A2+B2+2AB-B2AR=A

If A⇀+B⇀=R⇀ and 2A⇀+B⇀ is perpendicular to B⇀ then

A→+B→=R→ A→+B→=R given 2A→+B→ is perpendicular to B→ (2A→+B→)⋅B→=0⇒[A→+(A→+B→)]⋅B→=0(A→+R→)⋅B→=0⇒A→=−R→∴A=R [∵B≠0 and R→ cannot be perpendicular to B→]Alternatively 2A⇀+B →is perpendicular to B cosθ=-B2AR2=A2+B2+2AB-B2AR=A

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If $\overset{⇀}{A} + \overset{⇀}{B} = \overset{⇀}{R}$ and $2 \overset{⇀}{A} + \overset{⇀}{B}$ is perpendicular to $\overset{⇀}{B}$ then

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A = R

B = A

B = R

B = 2R

answer is A.

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Detailed Solution

$\vec{A} + \vec{B} = \vec{R}$

$\vec{A} + \vec{B} = R given 2 \vec{A} + \vec{B} is perpendicular to \vec{B}$

$(2 \vec{A} + \vec{B}) \cdot \vec{B} = 0$

$\Rightarrow [\vec{A} + (\vec{A} + \vec{B})] \cdot \vec{B} = 0$
$(\vec{A} + \vec{R}) \cdot \vec{B} = 0 \Rightarrow \vec{A} = - \vec{R} ∴ A = R$
$[∵ B \neq 0 and \vec{R} cannot be perpendicular to \vec{B}]$