Find the angle between the vectors A=2 hat{i} | vedclass

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(D) Given,$A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$.Let $	heta$ be the angle between $A$ and $B$.By using the dot produ

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$90$

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(D) Given,$A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$.Let $	heta$ be the angle between $A$ and $B$.By using the dot product formula,$A \cdot B = |A| |B| \cos 	heta$,where $|A|$ and $|B|$ are the magnitudes of vectors $A$ and $B$ respectively.First,calculate the magnitudes:$|A| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$|B| = \sqrt{4^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.Now,calculate the dot product:$A \cdot B = (2)(4) + (4)(2) + (4)(-4) = 8 + 8 - 16 = 0$.Substituting these values into the dot product formula:$0 = (6)(6) \cos 	heta$.$0 = 36 \cos 	heta$.$\cos 	heta = 0$.Therefore,$	heta = 90^{\circ}$.

Find the angle between the vectors $A=2 \hat{i}+4 \hat{j}+4 \hat{k}$ and $B=4 \hat{i}+2 \hat{j}-4 \hat{k}$. (in $^{\circ}$)

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