Vector addition, how can this vector be u-p?

Question

How can this vector that is orthogonal to p be u-p? it does not even have the same direction as p. (p is the projection of u on p)

Answer 1

$d = u-p$ must fulfill: \begin{align} p + d &= u \iff \\ p + (u-p) &= u \end{align} and this it does.

Answer 2

The projection is calculated as

$$ {\bf p} = ({\bf u}\cdot \hat{{\bf v}})\hat{{\bf v}} $$

Now calculate

$$ {\bf p}\cdot ({\bf u} - {\bf p}) = {\bf p}\cdot {\bf u} - {\bf p}\cdot {\bf p} = ({\bf u}\cdot \hat{{\bf v}})({\bf u}\cdot \hat{{\bf v}}) - ({\bf u}\cdot \hat{{\bf v}})({\bf u}\cdot \hat{{\bf v}}) = 0 $$

The vectors are orthogonal!

Answer 3

Imagine the vectors as instructions to walk a certain distance in a specified direction.

$p$ is walk east $4m$.

$u - p$ is walk north $3m$.

$u$ is walk approximately east north east $5m$.

Think of a $3, 4, 5$ triangle. You can get from the bottom left by the $5$ side or the $3$ and $4$ sides in sequence. The net effect is the same.