Let vec{a} = hat{i} + hat{j} + hat{k},vec{c} | vedclass

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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,so $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Given $\vec{c} = \hat{j} - \hat{k}$,so $|\vec{c}|

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$\sqrt{\frac{11}{3}}$

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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,so $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Given $\vec{c} = \hat{j} - \hat{k}$,so $|\vec{c}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.We have $\vec{a} 	imes \vec{b} = \vec{c}$. Taking the magnitude on both sides,$|\vec{a} 	imes \vec{b}| = |\vec{c}|$.$|\vec{a}| |\vec{b}| \sin 	heta = |\vec{c}|$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.$\sqrt{3} |\vec{b}| \sin 	heta = \sqrt{2} \quad \dots (1)$Also,$\vec{a} \cdot \vec{b} = 3$.$|\vec{a}| |\vec{b}| \cos 	heta = 3 \Rightarrow \sqrt{3} |\vec{b}| \cos 	heta = 3 \quad \dots (2)$Squaring and adding $(1)$ and $(2)$:$(\sqrt{3} |\vec{b}| \sin 	heta)^2 + (\sqrt{3} |\vec{b}| \cos 	heta)^2 = (\sqrt{2})^2 + 3^2$$3 |\vec{b}|^2 (\sin^2 	heta + \cos^2 	heta) = 2 + 9$$3 |\vec{b}|^2 = 11$$|\vec{b}|^2 = \frac{11}{3} \Rightarrow |\vec{b}| = \sqrt{\frac{11}{3}}$.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}|$ equals?

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