A number of properties follow immediately from the definition of the dot product, using the perpendicular vectors ${\bf i}\,, \ {\bf j}\,, \ {\bf k} $:

These properties provide a convenient algebraic way of computing the dot product of vectors $${\bf u}\, = \, \langle\,u_1,\, u_2,\, u_3\,\rangle\,=\,u_1\, {\bf i}+ u_2\,{\bf j}+ u_3\,{\bf k}, \qquad {\bf v}\,=\, \langle\,v_1,\, v_2,\, v_3\,\rangle\,=\,v_1\, {\bf i}+ v_2 \,{\bf j}+v_3\, {\bf k}\,.$$ By expanding using also Properties 1, 2, and 3, we get $${\bf u}\, \cdot \,{\bf v} \ = (u_1\, {\bf i}+ u_2\,{\bf j}+ u_3\,{\bf k})\,\cdot(\,v_1\, {\bf i}+ v_2 \,{\bf j}+v_3\, {\bf k})\ = \ \ u_1 v_1 + u_2 v_2 + u_3 v_3\,.$$

Example 1: Determine the dot product of the vectors $${\bf a} \ = \ \langle \, -3,\, 2,\, -3\,\rangle\,, \quad {\bf b} \ = \ \langle\, 2,\, 1,\, -3\,\rangle\,.$$ Solution: The dot product, ${\bf a}\,\cdot\,{\bf b} $, of vectors $${\bf a} \ = \ \langle \, a_1,,\, a_2,\, a_3\,\rangle\,, \quad {\bf b} \ = \ \langle\, b_1,\, b_2,\, b_3\,\rangle$$ is given by

$${\bf a}\cdot{\bf b} \ = \ a_1 b_1 + a_2 b_2 + a_3 b_3\,.$$ So when $${\bf a} \ = \ \langle \, -3,\, 2,\, -3\,\rangle\,, \quad {\bf b} \ = \ \langle\, 2,\, 1,\, -3\,\rangle$$ we see that $${\bf a}\,\cdot\,{\bf b} \ = \ (-3)(2)+(2)(1) + (-3)(-3) \ = \ 5 \, .$$

Example 2: determine the dot product of the vectors ${\bf a},\, {\bf b}$ when $$\|{\bf a}\| \ = \ 4,\qquad \|{\bf b}\| \ = \ 5$$ and the angle between ${\bf a},\, {\bf b}$ is $\pi/3$. Solution: the dot product is defined in

coordinate-free form by $$ {\bf a}\cdot{\bf b} \ = \ \|{\bf a}\|\,\|{\bf b}\|\,\cos \theta$$ where $\theta$ is the angle between them. When $\|{\bf a}\| = 4\,, \ \|{\bf b}\| \ = \ 5$ and $\theta = \pi/3$, therefore, $${\bf a}\,\cdot\,{\bf b} \ = \ 20 \cos \frac{\pi}{3} \ = \ 10\,.$$

Example 3: Find the angle beween the vectors ${\bf a},\, {\bf b}$ when ${\bf a} = \langle 1,1,1 \rangle$ and ${\bf b} = \langle 1, -1, 1 \rangle$. Solution: Using coordinates, we see that $${\bf a} \cdot {\bf b} = 1(1)+1(-1)+1(1)=1.$$

Since $$\|{\bf a}\|=\|{\bf b}\|=\sqrt{3},$$ $$\cos(\theta) = \frac{{\bf a}\cdot{\bf b}}{\|{\bf a}\|\, \|{\bf b}\|} = \frac{1}{3},$$ so $$\theta = \cos^{-1}(\frac{1}{3}) \approx 70.5^\circ.$$