# 11b.5. Abstract vectors involving the dot and cross products

Sections 11b.1 to 11b.4 have primarily focused on formulas where we compute the dot and/or cross when vectors are defined in column vector or $\mathbf{ijk}$ form. For more abstract vectors, we will have to rely on the following formulas:

The first row is a result from $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \cos \theta$ and $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| \, |\mathbf{b}| \sin \theta$, given that the angle a vector $\mathbf{a}$ makes with itself is 0.

The second row states that the order of a dot product is not important, while reversing the order of the cross product can only be done if we add a negative sign. This can be checked from the definitions of how the dot and cross products are computed. The third row states that constants can be brought in and/or out of a dot and cross products, which make computations easier.

Finally, the last row states that we can do expansion and/or factorization for both the dot and cross product. For the cross product, since the order is important, do note that vectors that are in front has to stay in front and vice versa.

The use of these formulas are illustrated in the two examples (plus an additional one) below:

### Solution to examples

Area of triangle $OBC$
$= \frac{1}{2} | \mathbf{c} \times \mathbf{b} | \\ = \frac{1}{2} | (3 \mathbf{a} + \mathbf{b}) \times \mathbf{b} | \\ = \frac{1}{2} | 3 \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{b}| \\ = \frac{1}{2} | 3 \mathbf{a} \times \mathbf{b} + \mathbf{0}|$

Length of projection of $\overrightarrow{OC}$ onto $\overrightarrow{OB}$
$= (3 \mathbf{a} + \mathbf{b}) \cdot \hat{\mathbf{b}} \\ = 3 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \\ = 3 |\mathbf{a}| \, |\mathbf{b}| \cos \theta + | \mathbf{b} |^2 \\ = 3 (2) (1) \left ( \frac{1}{2} \right ) + 1^2$
units.

$\mathbf{a} \times \mathbf{b} = 2 \mathbf{b} \times \mathbf{c} \\ \mathbf{a} \times \mathbf{b} - 2 \mathbf{b} \times \mathbf{c} = \mathbf{0} \\ \mathbf{a} \times \mathbf{b} + 2\mathbf{c} \times \mathbf{b} = \mathbf{0} \\ (\mathbf{a}+2\mathbf{c}) \times \mathbf{b} = \mathbf{0}$
Hence $\mathbf{a}+2\mathbf{c}$ is parallel to $\mathbf{b}$ so
for some $k \in \mathbb{R}, k \neq 0$.