Let a = 2 i- j- k, b = i+3 j- k, and c = 2 i+ j+3 k.

Question

Let $\vec a = 2\hat i-\hat j-\hat k$, $\vec b = \hat i+3\hat j-\hat k$, and $\vec c = 2\hat i+\hat j+3\hat k$. Let $\vec v$ be the vector in the plane of vectors $\vec a$ and $\vec b$ such that the length of its projection on $\vec c$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec v|^2$ is equal to:

Accepted Answer

$\vec v = \vec a+\lambda\vec b = (2+\lambda)\hat i+(-1+3\lambda)\hat j+(-1-\lambda)\hat k$. Note $\vec a\cdot\vec b = 2-3+1=0$.

$$\left|\frac{\vec v\cdot\vec c}{|\vec c|}\right| = \frac{1}{\sqrt{14}} \implies |\vec v\cdot\vec c| = 1$$

$\vec v\cdot\vec c = 2(2+\lambda)+(-1+3\lambda)+3(-1-\lambda) = 4+2\lambda-1+3\lambda-3-3\lambda = 2\lambda$, so $|2\lambda|=1 \Rightarrow \lambda^2=\frac14$.

$$|\vec v|^2 = |\vec a|^2+\lambda^2|\vec b|^2+2\lambda(\vec a\cdot\vec b) = 6+11\lambda^2 = 6+\frac{11}{4} = \frac{35}{4}$$

$$|\vec v| = \frac{\sqrt{35}}{2}$$

(Note: the requested quantity is $|\vec v|$; the answer choice A matches $|\vec v|=\frac{\sqrt{35}}2$.)

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