## Section3.6Scalar Product Between Vectors

Scalar or dot product of two vectors results in a single real number. To be specific, consider two vectors $\vec A$ and $\vec B$ that have an angle $\theta$ between them as shown in Figure 3.6.1. We denote the scalar product by placing a dot between the abstract symbols $\vec A$ and $\vec B$ for the two vectors, as $\vec A\cdot\vec B\text{.}$

The result of scalar product between $\vec A$ and $\vec B$ can be given in two equivalent ways: in terms of components of the vectors or in terms of magnitudes and angle.

\begin{align} \amp \text{(1) }\vec A \cdot\vec B = A_x B_x + A_y B_y + A_z B_z,\label{eq-dot-product-from-components}\tag{3.6.1}\\ \amp \text{(2) } \vec A \cdot\vec B = A B \cos\theta,\label{eq-dot-product-from-mag-and-angle}\tag{3.6.2} \end{align}

where $A$ and $B$ are magnitudes of these vectors.

\begin{align*} \amp A = \sqrt{A_x^2 + A_y^2 + A_z^2}\\ \amp B = \sqrt{B_x^2 + B_y^2 + B_z^2} \end{align*}

An important use of these formulas to find the angle between two vectors, given their components. Equating these Eqs. (3.6.1) and (3.6.2), and solving for $\cos\theta$ we get

$$\cos\theta = \dfrac{A_x B_x + A_y B_y + A_z B_z}{A\, B}.\tag{3.6.3}$$

Since components are simple numbers, their multiplication order does not matter, i.e., $A_xB_x = B_xA_x\text{,}$ therefore, the order of vectors in the dot product does not matter. That is,

$$\vec A \cdot \vec B = \vec B \cdot \vec A.\tag{3.6.4}$$

This will not be the case with the other product between vectors, called cross product, we will define below - we will find that reversing the order introduces a negative sign.

Consider the following vectors given in their component forms in a three-dimensional space, $\vec F = (3 \text{ N}, 4 \text{ N}, -5 \text{ N})\text{,}$ and $\vec d = (-4 \text{ m}, 5 \text{ m}, 3 \text{ m})\text{.}$ (a) Find the magnitudes of the two vectors. (b) Find the dot product of the two vectors. (c) What is the angle betwen the two vectors?

Hint

(a) Use $F = \sqrt{F_x^2 + F_y^2 + F_z^2}\text{,}$ (b) Use dot product based on components, (c) Use dot product based on magnitude.

(a) $7.07\text{ N} \text{,}$ $7.07\text{ m} \text{,}$ (b) $- 7\text{ N.m} \text{,}$ (c) $98^{\circ}$

Solution 1 (a)

(a) Using $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$ on $\vec F$ and similar formula on $\vec d$ we immediately get the answer. The only thing we need to notice that we will also get units.

\begin{align*} F \amp = \sqrt{F_x^2 + F_y^2 + F_z^2} \\ \amp = \sqrt{(3 \text{ N})^2 + (4 \text{ N})^2 + (-5 \text{ N})^2} = 7.07\text{ N}. \end{align*}

Similarly, $d = 7.07\text{ m}\text{.}$

Solution 2 (b)

(b) Since, vectors to be multiplied are given in the component forms, we use the definition of the scalar product based on the components.

\begin{align*} \vec F \cdot \vec d \amp = F_x d_x + F_y d_y + F_z d_z\\ \amp = (3 \text{ N})\times(-4 \text{ m}) + (4 \text{ N})\times(5 \text{ m}) + (-5 \text{ N})\times(3 \text{ m}) = - 7 \text{ N.m}. \end{align*}
Solution 3 (c)

(c) Since two definitions of scalar product would give the same value and one of them has the angle $\theta$ we seek. Therefore, we solve for $\theta$ to get

\begin{equation*} \cos\theta = \dfrac{\vec F \cdot \vec d}{ F \, d}. \end{equation*}

Now, using the numerical values, we find that

\begin{equation*} \cos\theta = \dfrac{- 7 \text{ N.m}}{ 7.07\text{ N} \times 7.07\text{ m}} = -0.14. \end{equation*}

Inverting this we get

\begin{equation*} \theta = 1.71 \text{ rad or } 98^{\circ}. \end{equation*}

When will the dot product between two non-zero vectors be zero?

Hint

Dot product is zero when vectors are perpendicular to each other, i.e., $\theta = 90^{\circ}\text{.}$

Solution

Let $A$ and $B$ be the magnitudes of the vectors and $\theta$ be the angle between them. Then, we want

\begin{equation*} A\, B\, \cos\,\theta = 0 \text{ when } A\ne 0 \text{ and } B\ne 0. \end{equation*}

That means,

\begin{equation*} \cos\, \theta = 0. \end{equation*}

Since, $\theta$ can be at most $180^{\circ}\text{,}$ there is only one value at which $\cos\, \theta = 0\text{.}$ That is, when

\begin{equation*} \theta = 90^{\circ}. \end{equation*}

That is, the dot product is zero when the two vector are perpendicular.

Find the dot products:

(a) between $\vec A =$ ( $30$ m/s, due East) and $\vec B =$ ( $40$ m/s, due $60^{\circ}$ North of East),

(b) between $\vec F =$ ( $100$ N, due East) and $\vec d =$ ( $40$ m, due $120^{\circ}$ Counterclockwise from East, i.e., towards North and $30^{\circ}$ past North).

Hint

Use $A B \cos\, \theta$

(a) $600\text{ m}^2/\text{s}^2\text{,}$ (b) $- 2000\text{ N.m}\text{.}$