Given that the reflection of a vector $v=(3,0,0,4)$ across a plane $P=\{x:x^Tu=0\}$ is parallel to $w=(2,0,1,3)$. Find $u$
The reflction across a plane perpendicular to $u$ is given by the House holder matrix $$ H_u=I-2\frac{uu^T}{u^Tu} $$ Here I think it says $$ H_uv=\lambda w $$ Since $||H_uv||=||v||=\lambda||w||\implies \lambda =\frac{||v||}{||w||}$ $$ H_uv=v-2\frac{uu^T}{u^Tu}v=\frac{||v||}{||w||}w\\ \begin{bmatrix}3\\0\\0\\4\end{bmatrix}-2\frac{uu^T}{u^Tu}\begin{bmatrix}3\\0\\0\\4\end{bmatrix}=\frac{5}{\sqrt{14}}\begin{bmatrix}2\\0\\1\\3\end{bmatrix} $$ How do I find $u$ ?
Can I take $u=v-\frac{||v||}{||w||}w$ ?
Yes. Your solution is correct. You can make the substitution to convince yourself of this.
Alternatively, you can exploit geometrical intuition. First, normalize $v$ and $w$:
$$ v' = \frac{v}{\|v\|} \\ w' = \frac{w}{\|w\|}. $$
Now you can decompose $v'$ and $w'$ into the components parallel and orthogonal to the reflection plane. Note that the last component is parallel to $u$.
By geometry, the parallel components of $v'$ and $w'$ will be the same and the orthogonal components will have equal length and direction and opposite orientation. Therefore, we can recover the component orthogonal to the reflection plane, i.e. parallel to $u$, by subtracting $v'$ and $w'$ (since then the components parallel to the reflection plane cancel out).
We conclude that
$$ u = v' - w' = \frac{v}{\|v\|} - \frac{w}{\|w\|} $$
which is a scalar multiple of your result:
$$ \|v\| \left(\frac{v}{\|v\|} - \frac{w}{\|w\|}\right) = v-\frac{\|v\|}{\|w\|}w. $$