If vec{A} and vec{B} are two non-zero vectors | vedclass

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(C) Given: $|\vec{A} + \vec{B}| = \frac{1}{2} |\vec{A} - \vec{B}|$ and $|\vec{A}| = 2|\vec{B}|$.Squaring both sides of the first equation:$|\vec{A} +

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$\cos^{-1}(-3/4)$

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(C) Given: $|\vec{A} + \vec{B}| = \frac{1}{2} |\vec{A} - \vec{B}|$ and $|\vec{A}| = 2|\vec{B}|$.Squaring both sides of the first equation:$|\vec{A} + \vec{B}|^2 = \frac{1}{4} |\vec{A} - \vec{B}|^2$$A^2 + B^2 + 2AB \cos 	heta = \frac{1}{4} (A^2 + B^2 - 2AB \cos 	heta)$$4(A^2 + B^2 + 2AB \cos 	heta) = A^2 + B^2 - 2AB \cos 	heta$$4A^2 + 4B^2 + 8AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$$3A^2 + 3B^2 + 10AB \cos 	heta = 0$Substitute $A = 2B$ into the equation:$3(2B)^2 + 3B^2 + 10(2B)(B) \cos 	heta = 0$$3(4B^2) + 3B^2 + 20B^2 \cos 	heta = 0$$12B^2 + 3B^2 + 20B^2 \cos 	heta = 0$$15B^2 + 20B^2 \cos 	heta = 0$$20B^2 \cos 	heta = -15B^2$$\cos 	heta = -\frac{15}{20} = -\frac{3}{4}$$	heta = \cos^{-1}\left(-\frac{3}{4}ight)$

If $\vec{A}$ and $\vec{B}$ are two non-zero vectors such that $|\vec{A} + \vec{B}| = \frac{|\vec{A} - \vec{B}|}{2}$ and $|\vec{A}| = 2|\vec{B}|$,then the angle between $\vec{A}$ and $\vec{B}$ is

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