1.2 - The Scalar (Dot) Product

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Introduction

The scalar product (also called the dot product or inner product) is a way to multiply two vectors together to obtain a scalar (a number without direction). This operation is fundamental in physics for calculating work, projections, angles between vectors, and many other applications.

Geometric Definition of the Scalar Product:

The scalar product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is denoted by $\mathbf{a} \cdot \mathbf{b}$ and is defined as:

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

where $\theta$ is the angle between the two vectors when they are placed tail-to-tail, and $a$ and $b$ are the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$ respectively.

When the vectors are written in component form using a Cartesian coordinate system, the scalar product is simply the sum of the products of the components.
Scalar Product in Component Form (Cartesian Coordinates):

The scalar product can be written as the sum of the products of the vector components. For two vectors $\mathbf{a} = a_x\hat{x} + a_y\hat{y} + a_z\hat{z}$ and $\mathbf{b} = b_x\hat{x} + b_y\hat{y} + b_z\hat{z}$ the scalar product is:

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

or equivalently:

$$\mathbf{a} \cdot \mathbf{b} = \sum_{i=1}^3 a_i b_i.$$

where the summation is over the vector components.

Key Properties

Properties of the Scalar Product:
  • Commutative: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
  • Distributive: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$
  • Scalar multiplication: $(k\mathbf{a}) \cdot \mathbf{b} = k(\mathbf{a} \cdot \mathbf{b})$ for any scalar $k$
  • Dot product with itself: $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = a^2$. In other words, the magnitude of a vector is the square root of the dot product of the vector with itself: $|\mathbf{a}| = \sqrt{\mathbf{a} \cdot \mathbf{a}}$.
  • Orthogonal vectors: If $\mathbf{a} \cdot \mathbf{b} = 0$, then the vectors are perpendicular (orthogonal)

Geometric Interpretation

The scalar product has an important geometric meaning. The quantity $\mathbf{a} \cdot \mathbf{b}$ equals the magnitude of $\mathbf{a}$ times the component of $\mathbf{b}$ in the direction of $\mathbf{a}$ (or vice versa). This is equivalent to:

$$\mathbf{a} \cdot \mathbf{b} = a \left[b\cos\theta\right] = b \left[a\cos\theta\right]$$

where $b\cos\theta$ is the projection of $\mathbf{b}$ onto the direction of $\mathbf{a}$.

The value of the scalar product depends on the angle $\theta$ between the two vectors. This leads to several important special cases:

Special Cases of the Scalar Product:

In general, the scalar product ranges from $-ab$ (when vectors are opposite) to $+ab$ (when vectors are parallel in the same direction), with zero occurring when the vectors are perpendicular.

Example 1: Basic Calculation

Calculate the scalar product of $\mathbf{a} = 3\hat{x} + 4\hat{y}$ and $\mathbf{b} = 2\hat{x} - \hat{y} + 5\hat{z}$.

Solution:

Using the component form:

$$\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$
$$\mathbf{a} \cdot \mathbf{b} = (3)(2) + (4)(-1) + (0)(5) = 6 - 4 + 0 = 2$$
Example 2: Find the Angle Between Vectors

Find the angle between $\mathbf{u} = \hat{x} + 2\hat{y} + 2\hat{z}$ and $\mathbf{v} = 3\hat{x} + 4\hat{z}$.

Solution:

First, calculate the scalar product:

$$\mathbf{u} \cdot \mathbf{v} = (1)(3) + (2)(0) + (2)(4) = 3 + 0 + 8 = 11$$

Next, find the magnitudes:

$$|\mathbf{u}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$$
$$|\mathbf{v}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{25} = 5$$

Using the definition $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$:

$$11 = (3)(5)\cos\theta $$
$$\theta = \arccos\left(\frac{11}{15}\right) \approx 42.8°$$
Example 3: Orthogonal Vectors

Show that $\mathbf{a} = 2\hat{x} - \hat{y} + 3\hat{z}$ and $\mathbf{b} = \hat{x} + 5\hat{y} + \hat{z}$ are perpendicular.

Solution:

Calculate the scalar product:

$$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (-1)(5) + (3)(1) = 2 - 5 + 3 = 0$$

Since $\mathbf{a} \cdot \mathbf{b} = 0$, the vectors are orthogonal (perpendicular) to each other.

Dot Products of Orthonormal Basis Vectors

An orthonormal basis is a set of vectors that are perpendicular to each other and have magnitude 1. Consider, for example, the Cartesian basis vectors $\hat{x}$, $\hat{y}$, and $\hat{z}$. Because the basis vectors are unit vectors (i.e. normalized), the dot product of any basis vectors with itself is 1. Because the basis vectors are perpendicular to each other, the dot product of any two different basis vectors is 0. Therefore, we can organize the dot products between all pairs of Cartesian basis vectors as:

$$\hat{x} \cdot \hat{x} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\hat{x} \cdot \hat{y} = 0$$ $$\hat{y} \cdot \hat{y} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\hat{y} \cdot \hat{z} = 0$$ $$\hat{z} \cdot \hat{z} = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\hat{z} \cdot \hat{x} = 0$$

These results can be succintly summarized using the generalized unit vectors $\hat{e}_i$, where $\hat{e}_1 = \hat{x}$, $\hat{e}_2 = \hat{y}$, and $\hat{e}_3 = \hat{z}$. We also introduce the Kronecker delta symbol $\delta_{ij}$, which returns 1 if the indices are the same and 0 if they are different:

Kronecker Delta Symbol:
$$\delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}$$
Therefore, we can write the dot products of the Cartesian basis vectors as:
Dot Products of Orthonormal Basis Vectors:
$$\hat{e}_i \cdot \hat{e}_j = \delta_{ij} $$

Projections: Finding Vector Components

A frequent application of the scalar product is calculating vector components in a given orthonormal coordinate system. To find the $x$ component of a vector $\mathbf{v}$, for example, we take the dot product of $\mathbf{v}$ with the unit vector $\hat{x}$:

$$v_x = \mathbf{v} \cdot \hat{x}$$
Similarly, the $y$ and $z$ components are given by:
$$v_y = \mathbf{v} \cdot \hat{y}$$ $$v_z = \mathbf{v} \cdot \hat{z}$$

When we take the dot product of a vector with a unit vector that defines a coordinate axis, we say we are projecting the vector onto that axis. this terminilogy conjurs images of shining a light on the vector perpendicular to the axis to create a shadow or "projection" of the vector onto the axis.

This method works becuse the unit vectors form an orthonormal basis, meaning they are perpendicular to each other and have magnitude 1. Let's see why the above method works. We start by taking the dot product of our vector $\mathbf{v}$ with the unit vector $\hat{x}$ and substituting the components of $\mathbf{v}$:

\begin{aligned}\mathbf{v} \cdot \hat{x} &= (v_x \hat{x} + v_y \hat{y} + v_z \hat{z}) \cdot \hat{x} \\ &= (v_x \hat{x}) \cdot \hat{x} + (v_y \hat{y})\cdot \hat{x} + (v_z \hat{z})\cdot \hat{x} \\ &= v_x (\hat{x} \cdot \hat{x}) + v_y (\hat{y}\cdot \hat{x} )+ v_z (\hat{z}\cdot \hat{x}) \\ &= v_x (1 ) + v_y (0)+ v_z (0)\\ &= v_x \end{aligned}

We used the fact that the unit vectors are orthonormal, i.e. $\hat{x} \cdot \hat{x} =1$, $\hat{y} \cdot \hat{x} =0$, and $\hat{y} \cdot \hat{x} =0$,.

These results can be generalized using the $\hat{e}_i$ basis vectors. Let's calculate the component of $\mathbf{v}$ in the direction of $\hat{e}_i$:

\begin{aligned} \mathbf{v} \cdot \hat{e}_i &= \left[ \sum_j v_j \hat{e}_j \right] \cdot \hat{e}_i \\ &= \sum_j ( v_j \hat{e}_j ) \cdot \hat{e}_i\\ &= \sum_j v_j (\hat{e}_j \cdot \hat{e}_i)\\ &= \sum_j v_j \delta_{ij}\\ &= v_i \end{aligned}
This derivation depends on the following property of the Kronecker delta symbol:
$$\sum_j A_j \delta_{ij} = A_i.$$
Notice that summing over the Kronecker delta symbol replaces $A_j$ inside the summation with the final result $A_i$.

Example 4: Dot Product of a Vector with Itself

Calculate $\mathbf{v} \cdot \mathbf{v}$ where $\mathbf{v} = 4\hat{x} - 3\hat{y}$.

Solution:

\begin{aligned} \mathbf{v} \cdot \mathbf{v} &= \sum_{i=1}^3 v_i v_i\\ &= (4)(4) + (-3)(-3) \\ &= 25. \end{aligned}

Note: this is the square of the magnitude of the vector, i.e. $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$.

Projecting a Vector onto an Arbitrary Axis

We can use the scalar product to project a vector onto an axis defined by any unit vector.

Projection of a vector onto an arbitrary axis

The projection of a vector $\mathbf{v}$ onto an arbitrary axis defined by a unit vector $\hat{n}$ is given by:

$$\mathbf{v} \cdot \hat{n}$$
This expression gives the component of $\mathbf{v}$ along $\hat{n}$.

Let's assume we know the components of each vector in Cartesian coordiantes given by
$$\mathbf{v} = \sum_i v_i \hat{e}_i $$
and
$$\hat{n} = \sum_i n_i \hat{e}_i $$
The component of $\mathbf{v}$ in the direction of $\hat{n}$ is given by:
$$v_n = \mathbf{v} \cdot \hat{n} = \sum_i v_i n_i $$

Example 5

Calculate the component of vector $\mathbf{r}$ along an axis defined by the unit vector $\hat{n}$, where

$$\mathbf{r} = 3\hat{x} + 4\hat{y} + 5\hat{z} $$ $$\hat{n} = \frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y}$$
Solution: We take the dot product of $\mathbf{r}$ with $\hat{n}$:
\begin{aligned} \mathbf{r} \cdot \hat{n} &= \sum_i v_i n_i\\ &= 3 \frac{\sqrt{3}}{2} + 4 \frac{1}{2} + 5 \cdot 0\\ &= \frac{3\sqrt{3} + 4}{2} \approx 4.60\\ \end{aligned}

Applications in Physics

The scalar product appears frequently in physics: