Mathematics · Ch 10 — Vector Algebra

Scalar (or Dot) Product of Two Vectors

10.6.1

Scalar (or Dot) Product of Two Vectors

10.6.1 Scalar (or Dot) Product of Two Vectors

The Fundamental Idea

The scalar product (also called the dot product) produces a real number — a scalar — from two vectors. It measures how much of one vector points in the direction of the other.

For two nonzero vectors $\vec{a}$ and $\vec{b}$ , with an angle $\theta$ between them (where $0 \leq \theta \leq \pi$ ), the scalar product is defined as:

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\,\cos\theta

If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ , the angle $\theta$ is not defined, and we define $\vec{a} \cdot \vec{b} = 0$ . The result is always a real number — never a vector.

Key Observations

Observation 1: The Result is a Real Number

$\vec{a} \cdot \vec{b}$ is always a real number (a scalar), whether the vectors are nonzero or zero vectors.

Observation 2: Perpendicular Vectors

Let $\vec{a}$ and $\vec{b}$ be two nonzero vectors. Then:

\vec{a} \cdot \vec{b} = 0 \quad \text{if and only if} \quad \vec{a} \perp \vec{b}

Why? When $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , so $|\vec{a}||\vec{b}|\cos 90^\circ = 0$ . Conversely, if $\vec{a} \cdot \vec{b} = 0$ with both vectors nonzero, then $\cos\theta = 0$ , which forces $\theta = 90^\circ$ .

Important

The dot product being zero is the algebraic test for perpendicularity between two nonzero vectors.

Observation 3: Parallel Vectors (Same Direction)

If $\theta = 0$ , then $\cos 0 = 1$ , so:

\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|

In particular, for any vector $\vec{a}$ :

\vec{a} \cdot \vec{a} = |\vec{a}|^2

This gives a direct way to find the magnitude of a vector: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$ .

Observation 4: Opposite Direction Vectors

If $\theta = \pi$ , then $\cos\pi = -1$ , so:

\vec{a} \cdot \vec{b} = -|\vec{a}|\,|\vec{b}|

In particular:

\vec{a} \cdot (-\vec{a}) = -|\vec{a}|^2

Observation 5: Dot Products of Unit Vectors

For the mutually perpendicular unit vectors $\hat{i}$ , $\hat{j}$ , $\hat{k}$ along the $x$ , $y$ , $z$ axes (each of magnitude 1):

Same direction ( $\theta = 0$ ):

\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1

Perpendicular ( $\theta = 90^\circ$ ):

\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0

Note

These results are the foundation for computing dot products in component form.

Observation 6: Finding the Angle Between Two Vectors

The angle $\theta$ between two nonzero vectors is:

\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} \quad\Longrightarrow\quad \theta = \cos^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}\right)

Observation 7: Commutativity

The scalar product is commutative:

\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

This holds because multiplication of real numbers is commutative, and the angle between $\vec{a}$ and $\vec{b}$ is the same as between $\vec{b}$ and $\vec{a}$ .

Two Important Properties of the Scalar Product

Property 1: Distributivity of Scalar Product Over Addition

For any three vectors $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ :

\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}

›Proof

The projection of $\vec{b} + \vec{c}$ onto $\vec{a}$ equals the sum of the projections of $\vec{b}$ and $\vec{c}$ onto $\vec{a}$ . Since the dot product equals $|\vec{a}|$ times the projection of the second vector onto $\vec{a}$ , and vector addition is linear, the distributive property follows. It can also be verified algebraically using components (developed below).

Property 2: Scalar Multiplication

For any vectors $\vec{a}$ , $\vec{b}$ and any scalar $\lambda$ :

(\lambda\vec{a}) \cdot \vec{b} = \lambda(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (\lambda\vec{b})

›Proof

$(\lambda\vec{a}) \cdot \vec{b} = |\lambda\vec{a}|\,|\vec{b}|\,\cos\theta$ . Since $|\lambda\vec{a}| = |\lambda|\,|\vec{a}|$ , and $\lambda\vec{a}$ points the same way as $\vec{a}$ (if $\lambda > 0$ ) or opposite (if $\lambda < 0$ ), the angle between $\lambda\vec{a}$ and $\vec{b}$ is $\theta$ or $\pi - \theta$ .

For $\lambda > 0$ : $(\lambda\vec{a}) \cdot \vec{b} = \lambda|\vec{a}|\,|\vec{b}|\,\cos\theta = \lambda(\vec{a} \cdot \vec{b})$

For $\lambda < 0$ : $(\lambda\vec{a}) \cdot \vec{b} = |\lambda||\vec{a}|\,|\vec{b}|\,\cos(\pi - \theta) = -|\lambda||\vec{a}|\,|\vec{b}|\cos\theta = \lambda|\vec{a}|\,|\vec{b}|\,\cos\theta = \lambda(\vec{a} \cdot \vec{b})$

The same reasoning gives $\vec{a} \cdot (\lambda\vec{b}) = \lambda(\vec{a} \cdot \vec{b})$ .

Dot Product in Component Form

Let:

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, \qquad \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}

Using Property 1 (distributivity) and Property 2 (scalar multiplication), the product expands into nine terms: …

Definition 2Scalar (dot) product of two vectors

Definition

The scalar (or dot) product of two nonzero vectors $\vec{a}$ and $\vec{b}$ , denoted by $\vec{a} \cdot \vec{b}$ , is defined as:

\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \, \cos \theta

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ (with $0 \le \theta \le \pi$ ).

If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ , then $\theta$ is not defined. In this case, we define $\vec{a} \cdot \vec{b} = 0$ .

Key Observations

The result $\vec{a} \cdot \vec{b}$ is a real number (a scalar), not a vector.
For nonzero vectors: $\vec{a} \cdot \vec{b} = 0$ if and only if $\vec{a}$ and $\vec{b}$ are perpendicular ( $\theta = 90^\circ$ ).
If $\theta = 0$ , then $\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}|$ . In particular, $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ .
If $\theta = \pi$ , then $\vec{a} \cdot \vec{b} = -|\vec{a}| \, |\vec{b}|$ .
For the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$ $\hat{i}, \hat{j}, \hat{k}$ :
- $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$
- $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$
The angle between two nonzero vectors is given by $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \, |\vec{b}|}$ .
The scalar product is commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ .

Intuition …

Property 1

Let $\vec{a}$ and $\vec{b}$ be any two vectors, and let $l$ be any scalar. Then the scalar (dot) product satisfies:

(l\vec{a}) \cdot \vec{b} = l (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (l\vec{b})

This means a scalar can be pulled out of either vector before taking the dot product without changing the result. It is used when simplifying expressions involving scalar multiplication inside a dot product, especially when vectors are …

Property 2

Let $\vec{a}$ and $\vec{b}$ be any two vectors, and let $l$ be any scalar. Then the scalar (dot) product satisfies:

(l\vec{a}) \cdot \vec{b} = l (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (l\vec{b})

Figure 10.19Two vectors a and b drawn from a common point with the angle theta between them, used to define the scalar (dot) product.

Fig. 10.19 — Two vectors a and b drawn from a common point with the angle theta between them, used to define the scalar (dot) product.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Fig 10.19 is a simple but essential diagram: it shows two vectors, a and b, drawn from the same starting point. Vector a is a horizontal arrow pointing to the right. Vector b is an arrow pointing up and to the right. Between them, a thin slate-coloured arc marks the smaller angle, labelled $\theta$ . That is all the figure contains — no axes, no grid, no extra labels.

The entire purpose of this drawing is to give a visual definition of the angle between two vectors. In the scalar (dot) product, $\theta$ is not just any angle; it is the smaller of the two angles formed when the vectors are placed tail-to-tail. The figure makes this concrete: you see the two arrows, you see the arc, and you understand that $\theta$ is the measure of the turn needed to align the direction of a with the direction of b (or vice versa).

From this picture, the textbook defines the scalar product:

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

Here, $|\mathbf{a}|$ and $|\mathbf{b}|$ are the magnitudes (lengths) of the vectors, and $\theta$ is the angle between them, exactly as shown in Fig 10.19. The formula tells you that the dot product is a real number, not a vector. If $\theta = 0^\circ$ , the vectors point in the same direction and $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|$ . If $\theta = 90^\circ$ , the vectors are perpendicular and $\mathbf{a}\cdot\mathbf{b} = 0$ . If $\theta = 180^\circ$ , they point opposite and the dot product is $-|\mathbf{a}||\mathbf{b}|$ .

Watch out

A common mistake is to think $\theta$ can be any angle, like $200^\circ$ . The figure shows the smaller angle between the two directions, so $\theta$ is always between $0^\circ$ and $180^\circ$ (or $0$ and $\pi$ radians). For $200^\circ$ , the smaller angle is actually $160^\circ$ , and $\cos(160^\circ)$ is negative — that is the correct value to use.

The diagram also grounds the idea that the dot product is commutative: $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$ . Swapping the vectors does not change the angle $\theta$ between them, so the formula gives the same result. This is obvious from the picture — the arc between the two arrows is the same regardless of which vector you name first. …

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