3.12 Scalar (Dot) ProductHL ONLY

1. Definitions of the Scalar Product

The scalar product (dot product) evaluates the relationship between two vectors, producing a scalar value rather than a vector.

Geometric Definition: For two non-zero vectors $\vec{u}$ and $\vec{v}$ with an angle $\theta$ between them ($0^\circ \le \theta \le 180^\circ$), the dot product is defined as:

$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta$

The sign of the dot product indicates the type of angle:

$\vec{u} \cdot \vec{v} > 0 \iff \theta$ is acute ($0^\circ \le \theta < 90^\circ$)
$\vec{u} \cdot \vec{v} = 0 \iff \theta$ is a right angle ($\theta = 90^\circ$)
$\vec{u} \cdot \vec{v} < 0 \iff \theta$ is obtuse ($90^\circ < \theta \le 180^\circ$)

Algebraic Definition: For vectors $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix}$, the dot product is the sum of the products of their components:

$\vec{u} \cdot \vec{v} = a_1 a_2 + b_1 b_2$

In 3D space, where $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}$, this expands to: $\vec{u} \cdot \vec{v} = a_1 a_2 + b_1 b_2 + c_1 c_2$

EXAMPLE 1 (Finding the Angle Between Vectors)

Given vectors $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$, find their magnitudes, dot product, and the angle $\theta$ between them.

Step 1: Find magnitudes
$|\vec{u}| = \sqrt{3^2 + 4^2} = 5$
$|\vec{v}| = \sqrt{1^2 + (-2)^2} = \sqrt{5}$

Step 2: Find the dot product
$\vec{u} \cdot \vec{v} = 3(1) + 4(-2) = 3 - 8 = \mathbf{-5}$

Step 3: Find the angle $\theta$

$$\begin{aligned} \cos\theta &= \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \\ &= \dfrac{-5}{5\sqrt{5}} = -\dfrac{1}{\sqrt{5}} \\ \theta &= \arccos\left(-\dfrac{1}{\sqrt{5}}\right) \approx \mathbf{116.6^\circ} \end{aligned}$$

2. Perpendicular and Parallel Conditions

Perpendicular (Orthogonal) Vectors: If two non-zero vectors are perpendicular, then $\theta = 90^\circ$ and $\cos 90^\circ = 0$. Therefore: $\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0$
Parallel Vectors: Two vectors are parallel if they are scalar multiples of each other ($\vec{u} = k\vec{v}$). In 2D, their components are proportional: $\vec{u} \parallel \vec{v} \iff \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = k$

EXAMPLE 2 (Finding a Perpendicular Vector)

To find a vector perpendicular to $\vec{u} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$, swap the components and change the sign of one component.

Choosing $\vec{v} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ yields:

$\vec{u} \cdot \vec{v} = 2(5) + 5(-2) = 10 - 10 = 0 \implies \vec{u} \perp \vec{v}$

Any scalar multiple, such as $\begin{pmatrix} -5 \\ 2 \end{pmatrix}$, is also perpendicular.

EXAMPLE 3 (Finding an Unknown Parameter)

Let $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} x \\ -6 \end{pmatrix}$. Find $x$ for each condition.

Condition A: Perpendicular vectors ($\vec{u} \perp \vec{v}$)

$$\begin{aligned} \vec{u} \cdot \vec{v} = 0 &\implies 3(x) + 4(-6) = 0 \\ &\implies 3x - 24 = 0 \\ &\implies 3x = 24 \implies \mathbf{x = 8} \end{aligned}$$

Condition B: Parallel vectors ($\vec{u} \parallel \vec{v}$)

$$\begin{aligned} \dfrac{x}{3} = \dfrac{-6}{4} &\implies 4x = -18 \\ &\implies x = -\dfrac{18}{4} \implies \mathbf{x = -4.5} \end{aligned}$$

3. Key Algebraic Properties

The dot product is commutative ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$) and distributive ($\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$). An important identity occurs when a vector is multiplied by itself:

$\vec{u} \cdot \vec{u} = |\vec{u}|^2 \quad \text{or} \quad \vec{u}^2 = |\vec{u}|^2$

Since the angle between a vector and itself is $0^\circ$ ($\cos 0^\circ = 1$), the dot product equals the square of its magnitude. This identity allows you to rewrite magnitudes as dot products.

EXAMPLE 4 (Geometric Proof Using Vectors)

If $|\vec{u} + \vec{v}| = |\vec{u} - \vec{v}|$ for two non-zero vectors, prove that $\vec{u}$ and $\vec{v}$ are perpendicular.

Proof:

$$\begin{aligned} |\vec{u} + \vec{v}|^2 &= |\vec{u} - \vec{v}|^2 && \text{(Square both sides)} \\ (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) &= (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) && \text{(Apply the magnitude identity)} \\ \vec{u}^2 + 2\vec{u} \cdot \vec{v} + \vec{v}^2 &= \vec{u}^2 - 2\vec{u} \cdot \vec{v} + \vec{v}^2 && \text{(Expand using the distributive law)} \\ |\vec{u}|^2 + 2\vec{u} \cdot \vec{v} + |\vec{v}|^2 &= |\vec{u}|^2 - 2\vec{u} \cdot \vec{v} + |\vec{v}|^2 && \text{(Substitute magnitudes)} \\ 4\vec{u} \cdot \vec{v} &= 0 && \text{(Simplify terms)} \\ \vec{u} \cdot \vec{v} &= 0 && \text{(Divide out the constant)} \end{aligned}$$

Since the dot product is zero, $\vec{u} \cdot \vec{v} = 0 \implies \mathbf{\vec{u} \perp \vec{v}}$. ■