Scalar (Dot) Product in 3D

Given two 3D vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$, the scalar (dot) product is defined by either equivalent formula:

The algebraic form is for computing; the geometric form is for interpreting. Three immediate consequences:

1. Self-product gives squared magnitude: $\mathbf{u} \cdot \mathbf{u} = u_1^2 + u_2^2 + u_3^2 = |\mathbf{u}|^2$.
2. Angle formula: $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ — rearranged from the geometric form.
3. Perpendicularity test: non-zero $\mathbf{u}, \mathbf{v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$ (because $\cos 90° = 0$).

Properties (for any vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and scalar $k$):

• Commutative: $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$
• Distributive: $\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$
• Scalar multiple: $(k\mathbf{u}) \cdot \mathbf{v} = k(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot (k\mathbf{v})$
• Positive definiteness: $\mathbf{u} \cdot \mathbf{u} \geq 0$, with equality iff $\mathbf{u} = \mathbf{0}$

Definition: $\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 = |\mathbf{u}||\mathbf{v}|\cos\theta$ · Result is a scalar (number), not a vector · $\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2$ · Properties: commutative, distributive, $(k\mathbf{u}) \cdot \mathbf{v} = k(\mathbf{u} \cdot \mathbf{v})$

Pause — copy the dot product definition $\mathbf{u}\cdot\mathbf{v} = \sum u_i v_i = |\mathbf{u}||\mathbf{v}|\cos\theta$, the scalar-result reminder, and the properties (commutative, distributive) into your book.

We just saw the dot product: $\mathbf{u}\cdot\mathbf{v} = u_1 v_1+u_2 v_2+u_3 v_3 = |\mathbf{u}||\mathbf{v}|\cos\theta$, a scalar with $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2$. That raises a question: how do we use this to find the actual angle between two vectors? This card answers it → $\cos\theta = (\mathbf{u}\cdot\mathbf{v})/(|\mathbf{u}||\mathbf{v}|)$ with $\theta \in [0°,180°]$; the sign of the dot product classifies the angle as acute, right, or obtuse.

Combining the two formulas gives a direct expression for the angle:

The procedure is always three steps:

1. Compute $\mathbf{u} \cdot \mathbf{v}$ using components.
2. Compute $|\mathbf{u}|$ and $|\mathbf{v}|$ using $\sqrt{u_1^2 + u_2^2 + u_3^2}$.
3. Substitute and take $\cos^{-1}$ to find $\theta$.

Classification by sign — read it off the dot product alone:

• $\mathbf{u} \cdot \mathbf{v} > 0$: $\cos\theta > 0$, so $\theta \in [0°, 90°)$ — acute.
• $\mathbf{u} \cdot \mathbf{v} = 0$: $\cos\theta = 0$, so $\theta = 90°$ — right angle.
• $\mathbf{u} \cdot \mathbf{v} < 0$: $\cos\theta < 0$, so $\theta \in (90°, 180°]$ — obtuse.

Angle formula: $\cos\theta = \dfrac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$ · $\theta \in [0°, 180°]$ — always take the non-negative angle · Sign of dot product classifies: positive = acute, zero = right, negative = obtuse · Use unit vectors to skip the denominator: if $|\hat{\mathbf{u}}| = |\hat{\mathbf{v}}| = 1$ then $\cos\theta = \hat{\mathbf{u}} \cdot \hat{\mathbf{v}}$

Pause — copy the angle formula $\cos\theta = (\mathbf{u}\cdot\mathbf{v})/(|\mathbf{u}||\mathbf{v}|)$, the range $\theta \in [0°,180°]$, the sign classification (positive → acute, zero → right, negative → obtuse), and the unit-vector shortcut into your book.