Vector Algebra Part 3

Concept:The projection length of $\vec{v}$ on $\vec{c}$ is given by $\dfrac{|\vec{v}\cdot\vec{c}|}{|\vec{c}|}$.Formula:$$\text{Projection length}=\dfrac{|\vec{v}\cdot\vec{c}|}{|\vec{c}|}$$For the intended form, take $\vec{v}=\vec{a}+\lambda\vec{b}$.Solution:Given $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$.$|\vec{c}|=\sqrt{2^2+1^2+3^2}=\sqrt{14}$.The projection condition gives $\left|\dfrac{\vec{v}\cdot\vec{c}}{|\vec{c}|}\right|=\dfrac{1}{\sqrt{14}}$, so $|\vec{v}\cdot\vec{c}|=1$.Let $\vec{v}=\vec{a}+\lambda\vec{b}$.Then $\vec{v}=(2+\lambda)\hat{i}+(-1+3\lambda)\hat{j}+(-1-\lambda)\hat{k}$.Compute $\vec{v}\cdot\vec{c}=(2+\lambda)(2)+(-1+3\lambda)(1)+(-1-\lambda)(3)=2\lambda$.Thus $|2\lambda|=1$, giving $\lambda=\pm\frac{1}{2}$.Also $\vec{a}\cdot\vec{b}=2(1)+(-1)(3)+(-1)(-1)=0$.So $\vec{a}\perp\vec{b}$.$|\vec{a}|^2=2^2+(-1)^2+(-1)^2=6$ and $|\vec{b}|^2=1^2+3^2+(-1)^2=11$.Therefore $|\vec{v}|^2=|\vec{a}|^2+\lambda^2|\vec{b}|^2=6+\frac{1}{4}(11)=\frac{35}{4}$.Hence $|\vec{v}|=\sqrt{\frac{35}{4}}=\dfrac{\sqrt{35}}{2}$.As written, a general vector in the plane is $\alpha\vec{a}+\beta\vec{b}$; the condition fixes only $\beta=\pm\frac{1}{2}$, so $|\vec{v}|$ is not unique unless the intended form $\vec{v}=\vec{a}+\lambda\vec{b}$ is assumed.Answer:$\dfrac{\sqrt{35}}{2}$, i.e. Option A (under the intended form).