Dot Products and Orthogonal Projection

Length and angle from a dot product

For real vectors, $\mathbf u\cdot\mathbf v=\sum_i u_iv_i$ and $\|\mathbf v\|=\sqrt{\mathbf v\cdot\mathbf v}$ . Nonzero vectors satisfy $\mathbf u\cdot\mathbf v=\|\mathbf u\|\|\mathbf v\|\cos\theta$ . Orthogonal vectors have zero dot product. The zero vector is orthogonal to every vector, but has no defined direction or angle.

The projection of $\mathbf b$ onto the line through nonzero $\mathbf a$ is

\mathbf p=\frac{\mathbf a^T\mathbf b}{\mathbf a^T\mathbf a}\mathbf a.

The residual $\mathbf r=\mathbf b-\mathbf p$ satisfies $\mathbf a^T\mathbf r=0$ . This condition derives the formula: solve $\mathbf a^T(\mathbf b-c\mathbf a)=0$ for $c$ .

Worked example: closest point on a line

For $\mathbf a=(1,1)^T$ and $\mathbf b=(3,1)^T$ , the coefficient is $4/2=2$ . Thus $\mathbf p=(2,2)^T$ and $\mathbf r=(1,-1)^T$ . The residual is perpendicular to the line. Any other point $\mathbf p+t\mathbf a$ has squared distance $\|\mathbf r\|^2+t^2\|\mathbf a\|^2$ , so the projection is the closest point.

Explore

Projection and perpendicular residual

Try this. Set b = (3,1) and the angle to 45°: the projection is (2,2). Rotate the line to 0° and 90°. The residual remains perpendicular; at b = (0,0), all three vectors vanish.

Vector b · x3Vector b · y1Line angle (degrees)45

b = (3, 1). Line angle = 45°. Projection p = (2, 2); residual r = (1, -1). Distance to line = 1.41421. Unit line direction · residual = 0. b = p + r. A zero vector appears as a dot.

Rotate the target line and move the input vector. The projection stays on the line and the residual stays perpendicular. The displayed dot product should remain zero apart from floating-point rounding.

Worked example: a projection matrix

For the same line, $P=\mathbf a\mathbf a^T/(\mathbf a^T\mathbf a)=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ . It satisfies $P^T=P$ and $P^2=P$ . Projecting twice changes nothing after the first projection. The complementary matrix $I-P$ extracts the residual.

If $\mathbf a=0$ , the formula divides by zero. Projection onto the zero subspace is simply zero, but it must be defined separately from projection onto a direction.

Practice

Project $(2,3)^T$ onto the horizontal axis.
Find the angle between $(1,0)^T$ and $(1,1)^T$ .
Verify the length decomposition for the worked example.

Show worked solutions

Projection $(2,0)^T$ , residual $(0,3)^T$ .
$\cos\theta=1/\sqrt2$ , so $\theta=\pi/4$ .
$\|b\|^2=10$ , $\|p\|^2=8$ , $\|r\|^2=2$ , so $10=8+2$ .