If |A|=2, |B|=5 and |A times B|=8. If the ang | vedclass

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(A) Given: $|A|=2, |B|=5$ and $|A 	imes B|=8$.We know that the magnitude of the cross product is given by $|A 	imes B| = |A| |B| \sin 	heta$.Substi

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$6$

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(A) Given: $|A|=2, |B|=5$ and $|A 	imes B|=8$.We know that the magnitude of the cross product is given by $|A 	imes B| = |A| |B| \sin 	heta$.Substituting the values: $8 = (2)(5) \sin 	heta$.$8 = 10 \sin 	heta \implies \sin 	heta = 8/10 = 4/5$.Since the angle $	heta$ is acute,$\cos 	heta$ must be positive.Using the identity $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$.The dot product is given by $A \cdot B = |A| |B| \cos 	heta$.Substituting the values: $A \cdot B = (2)(5)(3/5) = 10 	imes (3/5) = 6$.

If $|A|=2, |B|=5$ and $|A \times B|=8$. If the angle between $A$ and $B$ is acute,then $A \cdot B$ is

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