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 (not a vector).

Geometric Definition: Given vectors $\vec{u}$ and $\vec{v}$ separated by an included angle $\theta$, the dot product computes as:
$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta$
Algebraic Definition: Given component vectors $\vec{u} = \binom{a_1}{b_1}$ and $\vec{v} = \binom{a_2}{b_2}$, the dot product computes algebraically as:
$\vec{u} \cdot \vec{v} = a_1 a_2 + b_1 b_2$
(For 3D vectors, this seamlessly expands to $a_1 a_2 + b_1 b_2 + c_1 c_2$).

EXAMPLE 1

Given vectors $\vec{u} = \binom{3}{4}$ and $\vec{v} = \binom{1}{-2}$. Evaluate magnitudes, dot product, and included angle.

Magnitudes: $|\vec{u}| = 5$ and $|\vec{v}| = \sqrt{5}$.
Dot Product: $\vec{u} \cdot \vec{v} = 3(1) + 4(-2) = \mathbf{-5}$.
To find the angle, equate the geometric and algebraic definitions: $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{-5}{5\sqrt{5}} = -\frac{1}{\sqrt{5}}$. Solving yields $\theta \approx \mathbf{116.6^\circ}$.

2. Perpendicular and Parallel Conditions

  • Perpendicular (Orthogonal) Vectors: If the intersection angle is $90^\circ$, then $\cos 90^\circ = 0$. Therefore, two non-zero vectors are perpendicular if and only if $\mathbf{\vec{u} \cdot \vec{v} = 0}$.
  • Parallel Vectors: Two vectors share parallel geometry if they act as scalar multiples of each other: $\mathbf{\vec{u} = k\vec{v}}$. In coordinate form, their component ratios must be perfectly equivalent ($\frac{a_1}{a_2} = \frac{b_1}{b_2}$).

EXAMPLE 2 & 3

Example 2: To find vectors perpendicular to $\vec{u} = \binom{2}{5}$, swap the coordinates and negate exactly one of them. Vectors like $\binom{5}{-2}$ or $\binom{-5}{2}$ yield a dot product of $10 - 10 = 0$. Any scalar multiple of these also remains perpendicular.
Example 3: Let $\vec{u} = \binom{3}{4}$ and $\vec{v} = \binom{x}{-6}$.
If perpendicular: $\vec{u} \cdot \vec{v} = 0 \Rightarrow 3x - 24 = 0 \Rightarrow \mathbf{x = 8}$.
If parallel: Proportions align $\frac{x}{3} = \frac{-6}{4} \Rightarrow 4x = -18 \Rightarrow \mathbf{x = -4.5}$.

3. Key Algebraic Properties

The dot product satisfies commutative ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$) and distributive laws. A highly critical property emerges when a vector is dotted inherently with itself:

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

Because the angle between a vector and itself is $0^\circ$ (and $\cos 0^\circ = 1$), the dot product collapses purely to the square of its magnitude. This algebraic mechanism efficiently removes magnitudes from complex vector equations.

EXAMPLE 4

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

Proof:

Square both geometric sides: $|\vec{u} + \vec{v}|^2 = |\vec{u} - \vec{v}|^2$.
Convert to dot products using the magnitude property: $(\vec{u} + \vec{v})^2 = (\vec{u} - \vec{v})^2$.
Expand algebraically: $\vec{u}^2 + 2\vec{u} \cdot \vec{v} + \vec{v}^2 = \vec{u}^2 - 2\vec{u} \cdot \vec{v} + \vec{v}^2$.
Cancel identical magnitude terms: $4\vec{u} \cdot \vec{v} = 0 \Rightarrow \vec{u} \cdot \vec{v} = 0$.
Since the dot product is zero, $\vec{u} \perp \vec{v}$.