If |A|=2 and |B|=4 and the angle between them | vedclass

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(D) The magnitude of the difference of two vectors $A$ and $B$ is given by the formula:$|A-B| = \sqrt{|A|^2 + |B|^2 - 2|A||B|\cos	heta}$Given $|A| =

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$2\sqrt{3}$

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(D) The magnitude of the difference of two vectors $A$ and $B$ is given by the formula:$|A-B| = \sqrt{|A|^2 + |B|^2 - 2|A||B|\cos	heta}$Given $|A| = 2$,$|B| = 4$,and $	heta = 60^{\circ}$.Substituting the values into the formula:$|A-B| = \sqrt{2^2 + 4^2 - 2(2)(4)\cos(60^{\circ})}$$|A-B| = \sqrt{4 + 16 - 16 	imes \frac{1}{2}}$$|A-B| = \sqrt{20 - 8}$$|A-B| = \sqrt{12}$$|A-B| = 2\sqrt{3}$

If $|A|=2$ and $|B|=4$ and the angle between them is $60^{\circ}$,then $|A-B|$ is

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