Let vec{a}=hat{i}+2 hat{j}+3 hat{k},vec{b}=3 | vedclass

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(D) Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ and $\vec{c} \perp \vec{b}$,we can write $\vec{c} = \lambda (\vec{b} 	imes (\vec{a} 	im

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$\sqrt{\frac{11}{6}}$

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(D) Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ and $\vec{c} \perp \vec{b}$,we can write $\vec{c} = \lambda (\vec{b} 	imes (\vec{a} 	imes \vec{b}))$. Using the vector triple product formula $\vec{b} 	imes (\vec{a} 	imes \vec{b}) = (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}$. Calculate $\vec{b} \cdot \vec{b} = 3^2 + 1^2 + (-1)^2 = 11$ and $\vec{a} \cdot \vec{b} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$. Thus,$\vec{c} = \lambda (11 \vec{a} - 2 \vec{b}) = \lambda (11(\hat{i} + 2\hat{j} + 3\hat{k}) - 2(3\hat{i} + \hat{j} - \hat{k})) = \lambda (5\hat{i} + 20\hat{j} + 35\hat{k}) = 5\lambda (\hat{i} + 4\hat{j} + 7\hat{k})$. Given $\vec{a} \cdot \vec{c} = 5$,we have $5\lambda (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\hat{i} + 4\hat{j} + 7\hat{k}) = 5$. $5\lambda (1 + 8 + 21) = 5 \implies 30\lambda = 1 \implies \lambda = \frac{1}{30}$. Therefore,$\vec{c} = \frac{1}{6} (\hat{i} + 4\hat{j} + 7\hat{k})$. $|\vec{c}| = \frac{1}{6} \sqrt{1^2 + 4^2 + 7^2} = \frac{1}{6} \sqrt{1 + 16 + 49} = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$.

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c}=5$,then $|\vec{c}|$ is equal to

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