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**Ackbeet** Think of a simple example: $\displaystyle V$ is the xy plane, and $\displaystyle U$ is the subspace consisting of all points on the line $\displaystyle y=x.$ You can show that it is a subspace, because it contains the origin, and is closed under scalar multiplication and vector addition.

Now then. If your vector $\displaystyle v\in U,$ then obviously the best approximation is $\displaystyle v$ itself, because then $\displaystyle 0\le\|v-u\|=0.$ If, on the other hand, $\displaystyle v\not\in U,$ then what is the vector in $\displaystyle U$ closest to $\displaystyle v?$ Well, it would be the point on the line that is closest to $\displaystyle v.$ And that point is the point you would get by dropping the perpendicular from $\displaystyle v$ to the line $\displaystyle y=x.$ That also turns out to be the orthogonal projection onto the subspace $\displaystyle U.$

The essential idea here is that the straight line (in this case, the perpendicular), is the shortest path between two points, at least in the Euclidean norm. And that "perpendicular" is equivalent to the orthogonal projection onto the subspace.

Make sense? Does this help?