The angle between two vectors given by 6hat i | vedclass

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(D) Let $\vec{A} = 6\hat{i} + 6\hat{j} - 3\hat{k}$ and $\vec{B} = 7\hat{i} + 4\hat{j} + 4\hat{k}$.The dot product $\vec{A} \cdot \vec{B} = (6)(7) + (6

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$\sin^{-1}\left(\frac{\sqrt{5}}{3}\right)$

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(D) Let $\vec{A} = 6\hat{i} + 6\hat{j} - 3\hat{k}$ and $\vec{B} = 7\hat{i} + 4\hat{j} + 4\hat{k}$.The dot product $\vec{A} \cdot \vec{B} = (6)(7) + (6)(4) + (-3)(4) = 42 + 24 - 12 = 54$.The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{6^2 + 6^2 + (-3)^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$.The magnitude of $\vec{B}$ is $|\vec{B}| = \sqrt{7^2 + 4^2 + 4^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9$.Using the formula $\cos 	heta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}$,we get $\cos 	heta = \frac{54}{9 	imes 9} = \frac{54}{81} = \frac{2}{3}$.Since $\cos 	heta = \frac{2}{3}$,we can find $\sin 	heta$ using $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (2/3)^2} = \sqrt{1 - 4/9} = \sqrt{5/9} = \frac{\sqrt{5}}{3}$.Therefore,$	heta = \sin^{-1}\left(\frac{\sqrt{5}}{3}ight)$.

Column-$I$	Column-$II$
$(a) \|\vec{A} \times \vec{B}\| = 0$	$(i) \theta = 30^{\circ}$
$(b) \|\vec{A} \times \vec{B}\| = 8$	$(ii) \theta = 45^{\circ}$
$(c) \|\vec{A} \times \vec{B}\| = 4$	$(iii) \theta = 90^{\circ}$
$(d) \|\vec{A} \times \vec{B}\| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$

The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is

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