3.12 Scalar (Dot) Product - Angle Between Vectors (HL)

1. Definitions of the Scalar Product

The scalar product (dot product) evaluates the directional correlation between two intersecting vectors, producing a pure numerical scalar value rather than a vector quantity.

Geometric Definition: Given two non-zero vectors $\vec{u}$ and $\vec{v}$ separated by an included angle $\theta$ (where $0^\circ \le \theta \le 180^\circ$), the dot product computes as:

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

The sign of the dot product offers crucial geometric insights about the angle:

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

Algebraic Definition: Given component 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 computes algebraically by summing component-wise products:

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

For 3-Dimensional space vectors, 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 seamlessly expands to: $\vec{u} \cdot \vec{v} = a_1 a_2 + b_1 b_2 + c_1 c_2$

EXAMPLE 1 (Evaluating the Included Angle)

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

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

Step 2: Compute Algebraic Dot Product
$\vec{u} \cdot \vec{v} = 3(1) + 4(-2) = 3 - 8 = \mathbf{-5}$

Step 3: Equate Definitions to Extract 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 intersect at exactly $90^\circ$, then $\cos 90^\circ = 0$. Therefore: $\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0$
Parallel Vectors: Two vectors share identical or completely inverted directional geometry $\iff$ they act as linear scalar multiples of each other ($\vec{u} = k\vec{v}$). In 2D coordinate component tracking, their ratios must perfectly align: $\vec{u} \parallel \vec{v} \iff \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = k$

EXAMPLE 2 (Constructing Orthogonal Coordinates)

To easily discover a normal vector perpendicular to a given vector $\vec{u} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$, swap the components and invert exactly one sign.

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 tracking multiple $k\begin{pmatrix} 5 \\ -2 \end{pmatrix}$, such as $\begin{pmatrix} -5 \\ 2 \end{pmatrix}$, remains strictly perpendicular.

EXAMPLE 3 (Solving Parameter Boundary Constraints)

Let $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} x \\ -6 \end{pmatrix}$. Calculate the parameter value $x$ for both structural constraints.

Condition A: If the vectors are perpendicular ($\vec{u} \perp \vec{v}$)

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

Condition B: If the vectors are parallel ($\vec{u} \parallel \vec{v}$)

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

3. Key Algebraic Properties

The dot product satisfies the commutative law ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$) and the distributive law ($\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$). A highly critical identity emerges when a vector forms a scalar product with itself:

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

Because the separation angle between any vector and itself is $0^\circ$ (and $\cos 0^\circ = 1$), the product collapses directly to the square of its absolute geometric length. This algebraic tool acts as a mechanism to systematically convert magnitudes into dot product equations.

EXAMPLE 4 (Vector Geometric Proof)

For two non-zero vectors, it holds true that $|\vec{u} + \vec{v}| = |\vec{u} - \vec{v}|$. Prove that $\vec{u}$ and $\vec{v}$ are strictly perpendicular to each other.

Proof Implementation:

$$\begin{aligned} |\vec{u} + \vec{v}|^2 &= |\vec{u} - \vec{v}|^2 && \text{(Square both geometric expressions)} \\ (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) &= (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) && \text{(Apply self dot product magnitude identity)} \\ \vec{u}^2 + 2\vec{u} \cdot \vec{v} + \vec{v}^2 &= \vec{u}^2 - 2\vec{u} \cdot \vec{v} + \vec{v}^2 && \text{(Expand via Distributive Law)} \\ |\vec{u}|^2 + 2\vec{u} \cdot \vec{v} + |\vec{v}|^2 &= |\vec{u}|^2 - 2\vec{u} \cdot \vec{v} + |\vec{v}|^2 && \text{(Substitute vector length equivalents)} \\ 4\vec{u} \cdot \vec{v} &= 0 && \text{(Collect terms and subtract constants)} \\ \vec{u} \cdot \vec{v} &= 0 && \text{(Divide out scalar constant multiplier)} \end{aligned}$$

Since the resolved scalar dot product evaluates strictly to zero, $\vec{u} \cdot \vec{v} = 0 \implies \mathbf{\vec{u} \perp \vec{v}}$. ■