Scalar Product and Planes | NST Mathematics

Definition of the Scalar Product

The scalar product (or dot product) of two vectors $\mathbf{a}$ and $\mathbf{b}$ is:

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

where $\theta \in [0, \pi]$ is the angle between the vectors.

$Two vectors a and b with angle θ between them, showing the dot product projection$

The result is a scalar (hence the name).

Properties

Algebraic properties:

Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
Distributive: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$
Self-product: $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$

Basis vector products: $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1$ $\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0$

Component Form

For $\mathbf{a} = (a_x, a_y, a_z)$ and $\mathbf{b} = (b_x, b_y, b_z)$ :

$\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z$

This provides a computationally convenient formula derived from the basis vector properties.

Geometric Applications

Angle Between Vectors

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

$Two vectors a and b meeting at an angle θ, with the derivation of the angle formula shown$

Perpendicular Vectors

Two non-zero vectors are perpendicular (orthogonal) if and only if:

$\mathbf{a} \cdot \mathbf{b} = 0$

Component Along a Direction

The component of $\mathbf{a}$ in the direction of a unit vector $\hat{\mathbf{n}}$ :

$a_{\parallel} = \mathbf{a} \cdot \hat{\mathbf{n}} = |\mathbf{a}|\cos\theta$

This is the projection of $\mathbf{a}$ onto the direction $\hat{\mathbf{n}}$ .

Component Perpendicular to a Direction

$a_{\perp} = |\mathbf{a}|\sin\theta = \sqrt{|\mathbf{a}|^2 - (\mathbf{a} \cdot \hat{\mathbf{n}})^2}$

Equation of a Plane

Through a Point with Given Normal

A plane through point $\mathbf{a}$ with unit normal $\hat{\mathbf{n}}$ consists of all points $\mathbf{r}$ satisfying:

$(\mathbf{r} - \mathbf{a}) \cdot \hat{\mathbf{n}} = 0$

$Plane with normal vector n and point a on the plane, showing how any point r satisfies (r-a)·n = 0$

Why this works: the vector $\mathbf{r} - \mathbf{a}$ lies in the plane, so it must be perpendicular to the normal.

Expanded Form

If $\hat{\mathbf{n}} = (l, m, n)$ where $l, m, n$ are direction cosines (cosines of angles with coordinate axes):

$lx + my + nz = p$

where $p = \mathbf{a} \cdot \hat{\mathbf{n}}$ is the perpendicular distance from the origin to the plane.

Through Three Points

For three points $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ (not collinear), the plane is:

$\mathbf{r} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a}) + \nu(\mathbf{c} - \mathbf{a})$

The normal vector is:

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

Distance from Point to Plane

For a plane $\hat{\mathbf{n}} \cdot \mathbf{r} = p$ and point $\mathbf{q}$ :

$d = |\hat{\mathbf{n}} \cdot \mathbf{q} - p|$

This is the signed distance (positive on one side, negative on the other).

Why the Scalar Product Matters

The scalar product measures alignment between vectors:

Maximum when vectors are parallel
Zero when perpendicular
Negative when pointing in opposite directions

This captures the geometry of angles without explicitly computing angles, making it essential for:

Work done by a force: $W = \mathbf{F} \cdot \mathbf{d}$
Power delivered: $P = \mathbf{F} \cdot \mathbf{v}$
Flux through a surface: involves dot products with normals