Dot Product

The dot product of vectors can be used to determine the angle between two vectors.

The relationship between the dot product and the angle $\theta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by:

$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta$

Where:

• $\mathbf{a} \cdot \mathbf{b}$ is the dot product of vectors $\mathbf{a}$ and $\mathbf{b}$.
• $|\mathbf{a}|$ is the magnitude (length) of vector $\mathbf{a}$.
• $|\mathbf{b}|$ is the magnitude (length) of vector $\mathbf{b}$.
• $\theta$ is the angle between the two vectors, where $0 \le \theta \le \pi$ (or $0^\circ \le \theta \le 180^\circ$).

To find the angle $\theta$, we can rearrange the formula:

$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$

$\theta = \cos^{-1}\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\right)$

Steps to find the angle between two vectors:

1. Calculate the dot product $\mathbf{a} \cdot \mathbf{b}$.
2. Calculate the magnitude of vector $\mathbf{a}$, $|\mathbf{a}|$.
3. Calculate the magnitude of vector $\mathbf{b}$, $|\mathbf{b}|$.
4. Substitute the values into the formula for $\cos\theta$.
5. Calculate $\theta$ using the inverse cosine function ($\cos^{-1}$).

Proof of the dot product by the cosine rule

Consider triangle AOB with vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, where $\mathbf{c} = \mathbf{b} - \mathbf{a}$ and $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

Applying the cosine rule to triangle AOB:

$|\mathbf{c}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

$|\mathbf{b} - \mathbf{a}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

Expanding the left side using the dot product property $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$:

$(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

$\mathbf{b} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

$|\mathbf{b}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{a}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

Simplifying:

$-2\mathbf{a} \cdot \mathbf{b} = -2|\mathbf{a}||\mathbf{b}|\cos\theta$

$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$

Example 1

Determine the angle between the two vectors $\mathbf{a} = (3, -4, -1)$ and $\mathbf{b} = (0, 5, 2)$.

Solution

$\mathbf{a} \cdot \mathbf{b} = (3)(0) + (-4)(5) + (-1)(2) = 0 - 20 - 2 = -22$

$|\mathbf{a}| = \sqrt{3^2 + (-4)^2 + (-1)^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$

$|\mathbf{b}| = \sqrt{0^2 + 5^2 + 2^2} = \sqrt{0 + 25 + 4} = \sqrt{29}$

$\cos\theta = \frac{-22}{\sqrt{26}\sqrt{29}} = \frac{-22}{\sqrt{754}} \approx -0.8012$

$\theta = \cos^{-1}(-0.8012) \approx 143.24^\circ$

Example 2

Given two vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = -\mathbf{i} + 5\mathbf{j} + \mathbf{k}$, find:

(a) $\mathbf{a} \cdot \mathbf{b}$

(b) The angle between $\mathbf{a}$ and $\mathbf{b}$.

Solution

(a) $\mathbf{a} \cdot \mathbf{b} = (1)(-1) + (2)(5) + (3)(1) = -1 + 10 + 3 = 12$

(b) $|\mathbf{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

$|\mathbf{b}| = \sqrt{(-1)^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27}$

$\cos\theta = \frac{12}{\sqrt{14}\sqrt{27}} = \frac{12}{\sqrt{378}} \approx 0.6172$

$\theta = \cos^{-1}(0.6172) \approx 51.88^\circ$

Example 3

Calculate the angles of a triangle EFG with vertices $E(2, -3, 5)$, $F(0, 1, 4)$, and $G(-2, 5, 2)$.

Solution

$\overrightarrow{EF} = (0-2)\mathbf{i} + (1-(-3))\mathbf{j} + (4-5)\mathbf{k} = -2\mathbf{i} + 4\mathbf{j} - \mathbf{k}$

$\overrightarrow{EG} = (-2-2)\mathbf{i} + (5-(-3))\mathbf{j} + (2-5)\mathbf{k} = -4\mathbf{i} + 8\mathbf{j} - 3\mathbf{k}$

$\overrightarrow{FE} = 2\mathbf{i} - 4\mathbf{j} + \mathbf{k}$

$\overrightarrow{FG} = (-2-0)\mathbf{i} + (5-1)\mathbf{j} + (2-4)\mathbf{k} = -2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$

$\overrightarrow{GE} = 4\mathbf{i} - 8\mathbf{j} + 3\mathbf{k}$

$\overrightarrow{GF} = 2\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$

Angle E: $\cos E = \frac{\overrightarrow{EF} \cdot \overrightarrow{EG}}{|\overrightarrow{EF}| |\overrightarrow{EG}|} = \frac{(-2)(-4) + (4)(8) + (-1)(-3)}{\sqrt{21}\sqrt{89}} = \frac{8 + 32 + 3}{\sqrt{1869}} = \frac{43}{\sqrt{1869}} \approx 0.9945$, so $E \approx 5.94^\circ$

Angle F: $\cos F = \frac{\overrightarrow{FE} \cdot \overrightarrow{FG}}{|\overrightarrow{FE}| |\overrightarrow{FG}|} = \frac{(2)(-2) + (-4)(4) + (1)(-2)}{\sqrt{21}\sqrt{24}} = \frac{-4 - 16 - 2}{\sqrt{504}} = \frac{-22}{\sqrt{504}} \approx -0.9800$, so $F \approx 168.52^\circ$

Angle G: $\cos G = \frac{\overrightarrow{GE} \cdot \overrightarrow{GF}}{|\overrightarrow{GE}| |\overrightarrow{GF}|} = \frac{(4)(2) + (-8)(-4) + (3)(2)}{\sqrt{89}\sqrt{24}} = \frac{8 + 32 + 6}{\sqrt{2136}} = \frac{46}{\sqrt{2136}} \approx 0.9945$, so $G \approx 5.55^\circ$

Therefore, the angles of triangle EFG are approximately $5.94^\circ$, $168.52^\circ$, and $5.55^\circ$. (Note: There's a small rounding error, and the angles should sum close to 180°.)

Given two vectors of the same dimensions, one vector can be projected onto another. There are two types of projections: scalar projection and vector projection. Scalar projection gives the magnitude of the projection, while vector projection gives the projection vector itself.

The scalar projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$ is given by:

$\text{Proj}_\mathbf{b} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$

The vector projection of vector $\mathbf{a}$ onto vector $\mathbf{b}$ is given by:

$\text{Proj}_\mathbf{b} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$

Suppose $\mathbf{a}$ and $\mathbf{b}$ are two vectors, and $\theta$ is the angle between them.

Figure 1.

Figure 2.

Scalar projection

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is:

$\text{Proj}_\mathbf{b} \mathbf{a} = |\mathbf{a}| \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$

Similarly, the scalar projection of $\mathbf{b}$ onto $\mathbf{a}$ is:

$\text{Proj}_\mathbf{a} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$

Vector projection

The vector projection of $\mathbf{a}$ onto $\mathbf{b}$ is:

$\text{Proj}_\mathbf{b} \mathbf{a} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}$

Similarly, the vector projection of $\mathbf{b}$ onto $\mathbf{a}$ is:

$\text{Proj}_\mathbf{a} \mathbf{b} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2}\right)\mathbf{a}$

Example 1

Given vectors $\mathbf{a} = (1, 0, -2)$, $\mathbf{b} = (2, -1, 1)$, $\mathbf{p} = (3, 2, -2)$, and $\mathbf{q} = (2, 4, 6)$:

(a) Determine the vector projection of $\mathbf{a}$ onto $\mathbf{b}$.

(b) Find the scalar projection of $\mathbf{p}$ onto $\mathbf{q}$.

Solution

(a) $\mathbf{a} \cdot \mathbf{b} = (1)(2) + (0)(-1) + (-2)(1) = 2 + 0 - 2 = 0$

$|\mathbf{b}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$

$\text{Proj}_\mathbf{b} \mathbf{a} = \frac{0}{6}\mathbf{b} = \mathbf{0}$ (The zero vector)

(b) $\mathbf{p} \cdot \mathbf{q} = (3)(2) + (2)(4) + (-2)(6) = 6 + 8 - 12 = 2$

$|\mathbf{q}| = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = 2\sqrt{14}$

$\text{Proj}_\mathbf{q} \mathbf{p} = \frac{2}{2\sqrt{14}} = \frac{1}{\sqrt{14}} = \frac{\sqrt{14}}{14}$

Example 2

Find the vector projection of $\mathbf{a} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}$ onto $\mathbf{b} = 4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}$.

Solution

$\mathbf{a} \cdot \mathbf{b} = (1)(4) + (-2)(-4) + (1)(7) = 4 + 8 + 7 = 19$

$|\mathbf{b}|^2 = 4^2 + (-4)^2 + 7^2 = 16 + 16 + 49 = 81$

$\text{Proj}_\mathbf{b} \mathbf{a} = \frac{19}{81}(4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}) = \frac{76}{81}\mathbf{i} - \frac{76}{81}\mathbf{j} + \frac{133}{81}\mathbf{k}$

Consider Figure 2.12, where $\mathbf{a}$ and $\mathbf{b}$ are position vectors, and $C$ is the angle between them. Let $\mathbf{c}$ be the vector representing the side opposite angle $C$, so $\mathbf{c} = \mathbf{b} - \mathbf{a}$.

We want to prove the cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$, where $a = |\mathbf{a}|$, $b = |\mathbf{b}|$, and $c = |\mathbf{c}|$.

We know that $\mathbf{c} = \mathbf{b} - \mathbf{a}$. Taking the dot product of $\mathbf{c}$ with itself:

$\mathbf{c} \cdot \mathbf{c} = (\mathbf{b} - \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a})$

Expanding using the distributive property of the dot product:

$|\mathbf{c}|^2 = \mathbf{b} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{a}$

Since the dot product is commutative ($\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$), and using the property $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$, we have:

$c^2 = b^2 - 2(\mathbf{a} \cdot \mathbf{b}) + a^2$

We also know that the dot product can be expressed in terms of the angle $C$ between the vectors: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos C = ab\cos C$.

Substituting this into the previous equation:

$c^2 = a^2 + b^2 - 2ab\cos C$

This is the cosine rule.