Problem trying to find distances between vector $u$ and subspace $S$

Question

Given: $$S=\left(\begin{pmatrix}1\\2\\0\end{pmatrix},\begin{pmatrix}5\\0\\2\end{pmatrix}\right),\quad u=\begin{pmatrix}3\\6\\12\end{pmatrix}$$ Find the vector of the vector space $S$ which is closer to the vector $u$ and find the minimal distance between $u$ and $S$.

I can't figure out the way to find the distances stated in the problem. I do not know what projections need to be done as to find the subspace's vector closer to the given vector $u$.

Could anyone tell me what procedure should I follow?

Answer 1

I'll elaborate a bit the answer of $@Thomas Bakx:

To find the distance from a vector $v$ to the plane spanned by $u_1, u_2$, you need to find the length of the projection of $v$ on the direction perpendicular to this plane. A vector that give the perpendicular direction is $u_1\times u_2$. The length of the projection of $v$ on the direction given by this vector is $$\frac{| v \cdot (u_1\times u_2) | }{\|u_1\times u_2\|}$$

You can notice that $ v \cdot (u_1\times u_2) =\det(v,u_1,u_2)$, so the numerator is the volume of the parallelepiped formed by $v,u_1,u_2$. The denominator is the area of the base, the parallelogram on $u_1$, $u_2$. So we have here a known formula for the height.

To get the closest point, it will be $$w= v- \frac{v\cdot (u_1\times u_2)}{\|u_1\times u_2\|^2}\, u_1\times u_2$$

Answer 2

Observe that the point $P$ in $S$ closest to $u$ must connect $u$ to $P$ by a line perpendicular to the plane $S$. Can you see this in a picture?

Proceeding in this manner, can you find a vector normal to the plane $S$? Call it $v$. Now we just need the line through $u$ going in the direction of $v$ to intersect the plane $S$, and that point of intersection will be exactly the desired $P$. This amounts to solving three equations with three unknowns.