If - A - B -3 A - B- then the value of - A - B - iA- A-BB- A 2- B 2- AB -31 - 2- A 2- B 2-3 AB 1 - 2D- A 2- B 2- AB 1 - 2

The correct option is D (A^2 + B^2 + AB )^1/2|A×B|=√(3) (A . B) ⇒ AB sin θ = √(3) AB cos θ ⇒tan θ = √(3) ∴ θ = 60^o Now |R| = |A+B| = √(A^2 + B^2 + 2AB cos ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If | A × B |=...

Question

If $| \to A \times \to B | = \sqrt{3} \to A . \to B,$ then the value of $| \to A + \to B |$ is

{(A^{2} + B^{2} + \frac{A B}{\sqrt{3}})}^{1 / 2}

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A + B

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{(A^{2} + B^{2} + \sqrt{3} A B)}^{1 / 2}

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{(A^{2} + B^{2} + A B)}^{1 / 2}

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Solution

The correct option is D ${(A^{2} + B^{2} + A B)}^{1 / 2}$
$| \to A \times \to B | = \sqrt{3} (\to A . \to B)$
$\Rightarrow A B sin θ = \sqrt{3} A B cos θ$ $\Rightarrow tan θ = \sqrt{3}$
$∴ θ = 60^{o}$
Now $| \to R | = | \to A + \to B | = \sqrt{A^{2} + B^{2} + 2 A B cos θ}$
$= \sqrt{A^{2} + B^{2} + 2 A B (\frac{1}{2})}$
$= (A^{2} + B^{2} + A B)^{1 / 2}$

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The correct option is D (A2+B2+AB)1/2|→A×→B|=√3(→A.→B) ⇒AB sin θ=√3AB cos θ ⇒tan θ=√3 ∴ θ=60o Now |→R|=|→A+→B|=√A2+B2+2AB cos θ =√A2+B2+2AB(12) =(A2+B2+AB)1/2

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