The angle between the vectors 3,i + j + 2,k a | vedclass

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(B) Let $\vec{a} = 3\,i + j + 2\,k$ and $\vec{b} = 2\,i - 2\,j + 4\,k$.The dot product is $\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 +

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$\sin^{-1} \frac{2}{\sqrt{7}}$

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(B) Let $\vec{a} = 3\,i + j + 2\,k$ and $\vec{b} = 2\,i - 2\,j + 4\,k$.The dot product is $\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$.The magnitudes are $|\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$ and $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.Using the formula $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$,we get $\cos 	heta = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{6}{\sqrt{14} \cdot \sqrt{6}} = \frac{6}{\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}} = \frac{\sqrt{3}}{\sqrt{7}}$.Since $\cos 	heta = \frac{\sqrt{3}}{\sqrt{7}}$,we have $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{3}{7} = \frac{4}{7}$.Thus,$\sin 	heta = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$,which implies $	heta = \sin^{-1} \left( \frac{2}{\sqrt{7}} ight)$.

The angle between the vectors $3\,i + j + 2\,k$ and $2\,i - 2\,j + 4\,k$ is

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