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

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(C) Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$, we can write $\vec{v} = x\vec{a} + y\vec{b} = x(2\hat{i}-\hat{j}-\hat{k}) + y(\hat{i

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$\frac{\sqrt{35}}{2}$

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(C) Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$, we can write $\vec{v} = x\vec{a} + y\vec{b} = x(2\hat{i}-\hat{j}-\hat{k}) + y(\hat{i}+3\hat{j}-\hat{k}) = (2x+y)\hat{i} + (3y-x)\hat{j} - (x+y)\hat{k}$.The projection of $\vec{v}$ on $\vec{c}$ is given by $\left|\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}\right| = \frac{1}{\sqrt{14}}$.First, calculate $|\vec{c}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4+1+9} = \sqrt{14}$.Now, $\vec{v} \cdot \vec{c} = (2x+y)(2) + (3y-x)(1) + (-x-y)(3) = 4x + 2y + 3y - x - 3x - 3y = 2y$.Thus, $\left|\frac{2y}{\sqrt{14}}\right| = \frac{1}{\sqrt{14}} \implies |2y| = 1 \implies y^2 = \frac{1}{4}$.The magnitude squared is $|\vec{v}|^2 = (2x+y)^2 + (3y-x)^2 + (x+y)^2 = (4x^2 + 4xy + y^2) + (9y^2 - 6xy + x^2) + (x^2 + 2xy + y^2) = 6x^2 + 11y^2$.Substituting $y^2 = \frac{1}{4}$, we get $|\vec{v}|^2 = 6x^2 + \frac{11}{4}$.Assuming the question implies a specific vector where $x=1$, $|\vec{v}| = \sqrt{6 + 2.75} = \sqrt{8.75} = \sqrt{\frac{35}{4}} = \frac{\sqrt{35}}{2}$.

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