Let vec{a}=3 hat{i}+hat{j}-hat{k} and vec{c}= | vedclass

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(C) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$.$|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$,

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$66$

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(C) Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}$.$|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$,so $|\vec{a}| = \sqrt{11}$.$|\vec{c}|^2 = 2^2 + (-3)^2 + 3^2 = 4 + 9 + 9 = 22$,so $|\vec{c}| = \sqrt{22}$.Since $\vec{a} = \vec{b} 	imes \vec{c}$,$\vec{a}$ is perpendicular to $\vec{c}$,so $\vec{a} \cdot \vec{c} = 0$.Also,$|\vec{a}| = |\vec{b} 	imes \vec{c}| = |\vec{b}||\vec{c}| \sin 	heta$,where $	heta$ is the angle between $\vec{b}$ and $\vec{c}$.$\sqrt{11} = \sqrt{50} \cdot \sqrt{22} \sin 	heta \Rightarrow \sqrt{11} = 5\sqrt{2} \cdot \sqrt{22} \sin 	heta = 5 \cdot \sqrt{44} \sin 	heta = 10\sqrt{11} \sin 	heta$.Thus,$\sin 	heta = \frac{1}{10}$.Then $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{1}{100} = \frac{99}{100}$,so $\cos 	heta = \frac{\sqrt{99}}{10}$.Now,$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}| \cos 	heta$.$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2(\sqrt{50})(\sqrt{22}) \left(\frac{\sqrt{99}}{10}ight) = 72 + 2(5\sqrt{2})(\sqrt{22}) \left(\frac{3\sqrt{11}}{10}ight) = 72 + 2(5 \cdot 2 \cdot \sqrt{11}) \left(\frac{3\sqrt{11}}{10}ight) = 72 + 20\sqrt{11} \cdot \frac{3\sqrt{11}}{10} = 72 + 2(3 \cdot 11) = 72 + 66 = 138$.Finally,$|72 - |\vec{b} + \vec{c}|^2| = |72 - 138| = |-66| = 66$.

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$,then $|72-| \vec{b}+\vec{c}|^2 |$ is equal to $..........$.

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