Calculate scalar products from Cartesian components

Calculate scalar products from Cartesian components

Multiply matching coordinates and add the results. Keep negative signs with their components. The answer is a scalar; when testing perpendicular directions, compare it with zero and check that neither direction vector is zero.

In an orthonormal basis, write $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$. Then

$$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta.$$

The basis self-products are $\mathbf i\cdot\mathbf i=\mathbf j\cdot\mathbf j=\mathbf k\cdot\mathbf k=1$. Distinct perpendicular basis directions give $\mathbf i\cdot\mathbf j=\mathbf j\cdot\mathbf k=\mathbf k\cdot\mathbf i=0$, and their reversed products are zero by commutativity. The source basis sketch labels the positive y-direction with another i; use j there, as required by the printed formulas.

Derive the component rule

The complete expansion explains why only matching components survive:

$$\begin{aligned} \mathbf a\cdot\mathbf b&=(a_1\mathbf i+a_2\mathbf j+a_3\mathbf k)\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1\mathbf i\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &\quad+a_2\mathbf j\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &\quad+a_3\mathbf k\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1b_1\mathbf i\cdot\mathbf i+a_1b_2\mathbf i\cdot\mathbf j+a_1b_3\mathbf i\cdot\mathbf k\\ &\quad+a_2b_1\mathbf j\cdot\mathbf i+a_2b_2\mathbf j\cdot\mathbf j+a_2b_3\mathbf j\cdot\mathbf k\\ &\quad+a_3b_1\mathbf k\cdot\mathbf i+a_3b_2\mathbf k\cdot\mathbf j+a_3b_3\mathbf k\cdot\mathbf k\\ &=a_1b_1(1)+0+0+0+a_2b_2(1)+0+0+0+a_3b_3(1)\\ &=a_1b_1+a_2b_2+a_3b_3. \end{aligned}$$

Example 9 keeps both methods: form the columns first, or distribute and use self-products. Self-Practice 2 checks signs in two component calculations. Example 10 uses a zero product to find a parameter. Example 11 combines $\cos(\pi/3)=1/2$ with $\mathbf b\perp(\mathbf a-\mathbf b)$ to prove a relation between magnitudes; it does not make the two original vectors parallel.