The Scalar Product and Projections

Starting point: what you already know

• Vectors in component form, magnitude $|\mathbf{a}|=\sqrt{x^2+y^2}$, and unit vectors (previous chapter).
• The cosine rule $c^2=a^2+b^2-2ab\cos C$.
• $\cos 90^{\circ }=0$, and $\cos \theta <0$ for obtuse angles.
• Perpendicular lines have gradients that multiply to $-1$.

Two ways to calculate it

The answer is a number (a scalar), not a vector. The two formulas agree; the cosine rule shows why.

Perpendicular vectors

Useful properties

• $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$
• $\mathbf{a}\cdot \mathbf{a}=|\mathbf{a}{|}^2$
• $\mathbf{a}\cdot (\mathbf{b}+\mathbf{c})=\mathbf{a}\cdot \mathbf{b}+\mathbf{a}\cdot \mathbf{c}$
• $(k\mathbf{a})\cdot \mathbf{b}=k(\mathbf{a}\cdot \mathbf{b})$
• $\mathbf{i}\cdot \mathbf{i}=\mathbf{j}\cdot \mathbf{j}=1$ and $\mathbf{i}\cdot \mathbf{j}=0$

Projections

The projection of $\mathbf{a}$ onto $\mathbf{b}$ is the "shadow" of $\mathbf{a}$ on the line of $\mathbf{b}$, when light shines at right angles to $\mathbf{b}$.

%%FIG:projection%%

Quick summary

• $\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2=|\mathbf{a}||\mathbf{b}|\cos \theta$, a number.
• Use it to find the angle between vectors.
• $\mathbf{a}\cdot \mathbf{b}=0$ means perpendicular. Negative means obtuse.
• $\mathbf{a}\cdot \mathbf{a}=|\mathbf{a}{|}^2$.
• Projection of $\mathbf{a}$ onto $\mathbf{b}$: $\left(\frac{\mathbf{a}\cdot \mathbf{b}}{\mathbf{b}\cdot \mathbf{b}}\right)\mathbf{b}$.

Quiz