If A- - B- is perpendicular to B- and A- - 2 B- is perpendicular to A- then

A⃗+B⃗ is ⊥ to B⃗. nbsp; ⇒ (A⃗+B⃗).B⃗ = 0 ⇒A⃗.B⃗ + B^2 = 0 ⇒ ABcosθ + B^2 = 0 ⇒ Acosθ + B = 0 — (1) A⃗+2B⃗ is ⊥ to A⃗. nbsp; ⇒ (A⃗+2B⃗).A⃗ = 0 ⇒ nbsp;A^2 ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If A⃗ + B⃗ is...

Question

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

A = 2 B

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A = \frac{B}{2}

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A = \frac{B}{\sqrt{2}}

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A = \sqrt{2} B

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Solution

The correct option is D $A = \sqrt{2} B$
$\to A + \to B$ is $⊥$ to $\to B$ .
$\Rightarrow (\to A + \to B) . \to B = 0$
$\Rightarrow \to A . \to B + B^{2} = 0$
$\Rightarrow A B cos θ + B^{2} = 0$
$\Rightarrow A cos θ + B = 0$ --- (1)

$\to A + 2 \to B$ is $⊥$ to $\to A$ .
$\Rightarrow (\to A + 2 \to B) . \to A = 0$
$\Rightarrow A^{2} + 2 A B cos θ = 0$
$\Rightarrow A + 2 B cos θ = 0$ --- (2)
Solving (1) and (2), we get,
$A = \sqrt{2} B$

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