V3 Scalar product

August 12, 2022 - 10:37am — AJ (e67821)

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What is a scalar product? What is a dot product? This is the result of multiplying the magnitudes of the components of two or more vectors. The result is not a vector, but a scalar (which is without direction).

There are two ways to multiply two vectors:

The scalar or dot product which gives a number;
The vector or cross product which gives a vector.

In this module we consider the scalar or dot product.

Definition

The scalar, or dot, product of two vectors $\vec{a}\left(a_{1},a_{2},a_{3}\right)$ and $\vec{b}\left(b_{1},b_{2},b_{3}\right)$ is a scalar, defined by: \[ \vec{a}\cdot\vec{b}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \] or geometrically, \[ \vec{a}\cdot\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\cos\theta \] where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Properties of the Scalar or Dot Product

If $\vec{a}$ and $\vec{b}$ are non-zero vectors and $\vec{a}$ is perpendicular11 Perpendicular means at right angles to. A right angle is $90^{\circ}=\pi/2.$ to $\vec{b}$ then $\vec{a}\cdot\vec{b}=0,$ since $\cos\left(\frac{\pi}{2}\right)=0$.
If $\vec{a}$ is parallel to $\vec{b}$ then the angle between the vectors is $0$ and $\vec{a}\cdot\vec{b}=\mid\vec{a}\mid\mid\vec{b}\mid$ as $\cos\left(0\right)=1$.
The dot product does not depend on the order of multiplication: \[ \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a} \]
In three dimensions with $\hat{i},$ $\hat{j}$ and $\hat{k}$ unit vectors along the $x$, $y$ and $z$ axes respectively, we have: \[\begin{align*} \vec{i}\cdot\vec{j} & =\vec{j}\cdot\vec{k}=\vec{k}\cdot\vec{i}=0\\ \vec{i}\cdot\vec{i} & =\vec{j}\cdot\vec{j}=\vec{k}\cdot\vec{k}=1 \end{align*}\]

Examples

$\left(2\vec{i}+3\vec{j}+4\vec{k}\right)\cdot\left(-\vec{i}-2\vec{j}+\vec{k}\right)=\left(2\times\left(-1\right)\right)+\left(3\times\left(-2\right)\right)+\left(4\times1\right)=-4$
$\left(2,-3,-3\right)\cdot\left(1,1,-2\right)=2-3+6=5$
$\left(5,0,-1\right)\cdot\left(1,4,3\right)=5+0-3=2$
$\left(2\vec{i}+4\vec{k}\right)\cdot\left(-3\vec{i}-2\vec{j}\right)=2\times\left(-3\right)+0\times\left(-2\right)+4\times0=-6$

See Exercises 1, 2, and 3.

Angle Between Two Vectors

The angle $\theta,\left(0\leq\theta\leq\pi\right)$, between two vectors can be found using the definition of the dot product: \[\begin{align*} \vec{a}\cdot\vec{b} & =\mid\vec{a}\mid\mid\vec{b}\mid\cos\theta. \end{align*}\] Rearranging, \[\begin{align*} \cos\theta & =\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid} \end{align*}\] and \[\begin{align*} \theta & =\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right). \end{align*}\]

Examples

If $\vec{a}=\left(2,3,1\right)$ and $\vec{b}=\left(5,-2,2\right)$ find the angle $\theta$, between $\vec{a}$ and $\vec{b}$ \[\begin{align*} \theta & =\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right)\\ \vec{a}\cdot\vec{b} & =\left(2,3,1\right)\cdot\left(5,-2,2\right)=6\\ \mid\vec{a}\mid & =\sqrt{2^{2}+3^{2}+1^{2}}=\sqrt{14},\mid\vec{b}\mid=\sqrt{25+4+4}=\sqrt{33}\\ \theta & =\cos^{-1}\left(\frac{6}{\sqrt{33}\times\sqrt{14}}\right)\\ & =\cos^{-1}\left(0.2791\right)\\ \theta & =73.8^{\circ}. \end{align*}\] The angle between $\vec{a}$ and $\vec{b}$ is $73.8^{\circ}$.
Find the angle $\theta$, between $\vec{a}\left(1,0,1\right)$ and $\vec{b}\left(-2,-1,1\right).$ \[\begin{align*} \vec{a}\cdot\vec{b} & =\left(1,0,1\right)\cdot\left(-2,-1,1\right)=-1\\ \mid\vec{a}\mid & =\sqrt{2}\\ \mid\vec{b}\mid & =\sqrt{6}\\ \theta & =\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{\mid\vec{a}\mid\mid\vec{b}\mid}\right)=\cos^{-1}\left(\frac{-1}{\sqrt{2}\times\sqrt{6}}\right)=\cos^{-1}\left(-0.2887\right)\\ \theta & =106.8^{\circ}. \end{align*}\]

The angle between $\vec{a}$ and $\vec{b}$ is $106.8^{\circ}$.

See Exercises 4 and 5.

Exercise 1

Calculate the dot product of:

$\text{(a) $\left(2,5,-1\right)$ and $\left(4,1,1\right)$ }$

$\text{(b) $3\vec{i}$ }$ and $5\vec{j}$

$\text{(c)}$ $5\vec{k}$ and $\left(\vec{j}+2\vec{k}\right)$

Answers

$\text{(a) 12}\quad\text{(b) $0\quad\text{(c) $10$ }$ }$

Exercise 2

Find:

$\text{(a)$\,\left(2,0,4\right)\cdot\left(-3,1,3\right)$ }$

$\text{(b)$\,\left(0,5,1\right)\cdot\left(4,0,0\right)$ }$

$\text{(c)$\,\left(2\vec{i}+3\vec{k}\right)\cdot\left(7\vec{i}+2\vec{j}+4\vec{k}\right)$ }$

Answers

$\text{(a) 6}\quad\text{(b) $0\quad\text{(c) $26$ }$ }$

Exercise 3

Which of the following vectors are perpendicular?

$\text{(a)}\,\left(5,2,3\right)$

$\text{(b)}\,\left(0,1,-1\right)$

$\text{(c)}\,\left(-2,2,2\right)$

Answers

$\text{(a) and (c), (b) and (c)}$

Exercise 4

Find the angle between the following pairs of vectors:

$\text{(a)}\,\left(1,2,3\right)\ $and $\left(4,-1,0\right)$

$\text{(b)}\,\left(2,1,-2\right)\ $and $\left(1,5,-1\right)$

$\text{(c)}\,\left(0,5,1\right)\ $and $\left(2,0,0\right)$

$\text{(d)}\,\left(1,-2,3\right)\ $and $\left(-4,1,-3\right)$

$\text{(e)}\,\left(2,1,-2\right)\ $and $\left(0,4,0\right)$

$\text{(f)}\,\left(0,3,0\right)\ $and $\left(0,1,0\right)$

Answers

$\text{(a)$\,82.6^{\circ}\quad\text{(b)$\,54.7^{\circ}\quad\text{(c)$\,90^{\circ}\quad\text{(d)$\,141.8^{\circ}\quad\text{(e)$\,$ $70.5^{\circ}\quad\text{(f)$\,0^{\circ}$ }$ }$ }$ }$ }$ }$

Exercise 5

If $\vec{a}=\left(2,2,2\right),\,\vec{b}=\left(3,2,-1\right),$ and $\vec{c}=\left(-1,4,1\right),$

$\text{(a)$\,$ }$Show $\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$

$\text{(b)$\,$ }$Rearranging $\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ gives $\vec{a}\cdot\left(\vec{b}-\vec{c}\right)=0$. As $\vec{b}\neq\vec{c}$ what is the relationship between $\vec{a}$ and $\left(\vec{b}-\vec{c}\right)$?