Scalar Product of Two Vectors

The scalar product (or dot product) of two vectors combines them to produce an ordinary number (a scalar), not another vector — hence the name. For vectors $\vec A$ and $\vec B$ with angle $\theta$ between them,

Properties:

1. $\vec A\cdot\vec B$ is always a scalar; it is positive when $\theta$ is acute ($<90°$) and negative when $\theta$ is obtuse ($90°<\theta<180°$).
2. It is commutative: $\vec A\cdot\vec B = \vec B\cdot\vec A$.
3. It is distributive over addition: $\vec A\cdot(\vec B+\vec C) = \vec A\cdot\vec B + \vec A\cdot\vec C$.
4. The angle between two vectors can be recovered from their dot product: $\theta = \cos^{-1}\!\left(\dfrac{\vec A\cdot\vec B}{AB}\right)$.
5. Maximum when $\cos\theta=1$, i.e. $\theta=0°$ (parallel vectors): $(\vec A\cdot\vec B)_{max} = AB$.
6. Minimum (most negative) when $\cos\theta=-1$, i.e. $\theta=180°$ (anti-parallel vectors): $(\vec A\cdot\vec B)_{min} = -AB$.
7. If $\vec A \perp \vec B$, then $\vec A\cdot\vec B = 0$ (since $\cos90°=0$) — this is exactly how orthogonality (perpendicularity) of two vectors is tested/deduced: compute the dot product; if it comes out to zero, the vectors are perpendicular.
8. Self-dot product: $\vec A\cdot\vec A = AA\cos0° = A^2$, so the magnitude can also be written $A = \sqrt{\vec A\cdot\vec A}$.
9. For any unit vector, $\hat n\cdot\hat n = 1\times1\times\cos0° = 1$; in particular $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$.
10. Since $\hat i,\hat j,\hat k$ are mutually perpendicular, $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$.
11. In component form:

(every cross term like $A_x\hat i\cdot B_y\hat j$ vanishes because $\hat i\cdot\hat j=0$). …