Properties of Scalar Product

1. Commutative. $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=|\vec b||\vec a|\cos\theta=\vec b\cdot\vec a$.
2. Sign follows the angle. Since $0\le\theta\le\pi$: $\vec a\cdot\vec b=|\vec a||\vec b|$ when $\theta=0$ (parallel, same direction); $\vec a\cdot\vec b=-|\vec a||\vec b|$ when $\theta=\pi$ (parallel, opposite direction); $\vec a\cdot\vec b=0$ when $\theta=\pi/2$ (perpendicular). More generally, $\vec a\cdot\vec b>0$ for $0\le\theta<\pi/2$, and $\vec a\cdot\vec b<0$ for $\pi/2<\theta\le\pi$. (iii)–(iv) The zero-product test. $\vec a\cdot\vec b=0\iff |\vec a|=0$ or $|\vec b|=0$ or $\theta=\pi/2$. So for non-zero vectors, $\vec a\cdot\vec b=0$ is exactly the condition $\vec a\perp\vec b$.

(v) Self dot-product. $\vec a\cdot\vec a=|\vec a|^2$, often abbreviated $a^2$ — this identity is used constantly when expanding expressions like $|\vec a\pm\vec b|^2$.

(vi) The axis unit vectors. $\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1$ (each makes angle $0$ with itself), while $\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0$ (each pair is mutually perpendicular).

(vii) Scalars pull straight out. For scalars $\lambda,\mu$: $(\lambda\vec a)\cdot(\mu\vec b)=(\lambda\mu)(\vec a\cdot\vec b)$.

(viii) Distributive. $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$ (left distributivity), and $(\vec a+\vec b)\cdot\vec c=\vec a\cdot\vec c+\vec b\cdot\vec c$ (right distributivity); the same holds with $-$ in place of $+$, and extends to sums of any number of vectors.

(ix) Vector identities (proved just like $(x\pm y)^2$, $(x+y)(x-y)$ for real numbers, using (v) and (viii)):

$|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\,\vec a\cdot\vec b,\qquad |\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\,\vec a\cdot\vec b,$ $(\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2.$ Proof of the first: $(\vec a+\vec b)\cdot(\vec a+\vec b)=\vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b$ (using commutativity to combine the two cross terms).

(x) Coordinate (working) formula. For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k,\ \vec b=b_1\hat i+b_2\hat j+b_3\hat k$, expanding $\vec a\cdot\vec b$ term by term and using (vi) to kill every cross term between different axis vectors leaves only the three matching terms: $\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3.$ In words: the dot product is the sum of the products of corresponding components.

(xi) Angle formula. Rearranging the definition: $\theta=\cos^{-1}\!\left(\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right).$ …