Vector Algebra Part 3

Concept:A vector in the plane of two given vectors can be expressed as a linear combination of them.The length of projection of a vector on another is the absolute value of their dot product divided by the magnitude of the second vector.Explanation:Given $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$, and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$.Since $\vec{v}$ lies in the plane of $\vec{a}$ and $\vec{b}$, write $\vec{v} = \vec{a} + \lambda \vec{b}$.So $\vec{v} = (2+\lambda)\hat{i} + (-1+3\lambda)\hat{j} + (-1-\lambda)\hat{k}$.The length of projection of $\vec{v}$ on $\vec{c}$ is $\left|\frac{\vec{v}\cdot\vec{c}}{|\vec{c}|}\right| = \frac{1}{\sqrt{14}}$.Compute $|\vec{c}| = \sqrt{2^2+1^2+3^2} = \sqrt{14}$.Thus $|\vec{v}\cdot\vec{c}| = \frac{1}{\sqrt{14}} \times \sqrt{14} = 1$.Now $\vec{v}\cdot\vec{c} = (2+\lambda)(2) + (-1+3\lambda)(1) + (-1-\lambda)(3)$.Simplify: $4+2\lambda -1+3\lambda -3-3\lambda = 2\lambda$.So $|2\lambda| = 1$, giving $\lambda = \pm\frac{1}{2}$, hence $\lambda^2 = \frac{1}{4}$.Compute $|\vec{v}|^2 = |\vec{a} + \lambda\vec{b}|^2 = |\vec{a}|^2 + \lambda^2|\vec{b}|^2 + 2\lambda(\vec{a}\cdot\vec{b})$.Check $\vec{a}\cdot\vec{b} = 2(1) + (-1)(3) + (-1)(-1) = 2-3+1=0$ (orthogonal).$|\vec{a}|^2 = 2^2+(-1)^2+(-1)^2 = 6$.$|\vec{b}|^2 = 1^2+3^2+(-1)^2 = 11$.So $|\vec{v}|^2 = 6 + \frac{1}{4}(11) + 0 = \frac{24+11}{4} = \frac{35}{4}$.Therefore $|\vec{v}| = \sqrt{\frac{35}{4}} = \frac{\sqrt{35}}{2}$.Answer:$\frac{\sqrt{35}}{2}$ (Option A).