If - A - B -3 A - B- then value of - A - B - isA- A2-B2-A B-31-2B- A-BC- A2-B2-A B 2D- A2-B2-3 A B 2

The correct option is C ( A^2+ B^2+ AB )^12|A×B| = √(3) A. B AB sinθ = √(3) AB cosθ ⇒tanθ = √(3) ∴θ = 60^∘ Now, |R| = |A + B| = √(A^2+ B^2+ 2AB cosθ) = √(A^2+ ...

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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 value of $| \to A + \to B |$ is

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

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A + B

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

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

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Solution

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

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If |→A×→B|=√3 →A.→B , then value of |→A+→B| is

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