First slide

Vectors

Question

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

Moderate

A

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

A+B
C

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

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

Solution

$| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} . \vec{B})$
$ABsinθ = \sqrt{3} ABcosθ \Rightarrow tanθ = \sqrt{3} ∴ θ = 60^{0}$
Now $| \vec{R} | = | \vec{A} + \vec{B} | = \sqrt{A^{2} + B^{2} + 2 ABcosθ}$
= $\sqrt{A^{2} + B^{2} + 2 AB (\frac{1}{2})} = {(A^{2} + B^{2} + AB)}^{\frac{1}{2}}$

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