Let $\vec{u},\vec{v},$ and $\vec{w}$ be vectors such that $|\vec{u}|=1,|\vec{v}|=2,$ and $|\vec{w}|=3$. If the projection of $\vec{v}$ along $\vec{u}$ is equal to the projection of $\vec{w}$ along $\vec{u}$, and $\vec{v}\perp \vec{w}$, then the magnitude of $\vec{u}-\vec{v}+\vec{w}$ is:

Solution

1. Let $$\begin{array}{rl}\vec{u}\cdot \vec{v}& =\vec{u}\cdot \vec{w}\\ & =k\end{array}$$.
2. The magnitude squared is $|\vec{u}-\vec{v}+\vec{w}{|}^2=|\vec{u}{|}^2+|\vec{v}{|}^2+|\vec{w}{|}^2-2\vec{u}\cdot \vec{v}+2\vec{u}\cdot \vec{w}-2\vec{v}\cdot \vec{w}$.
3. Given $$\begin{array}{rl}|\vec{u}|& =1,|\vec{v}|\\ & =2,|\vec{w}|\\ & =3\end{array}$$ and $\vec{v}\cdot \vec{w}=0$, the expression simplifies to $1 + 4 + 9 - 2k + 2k - 0 = 14$.
4. Hence the answer is (C).