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 perpendicular1 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: \[\begin{align*} \vec{i}\cdot\vec{j} & =\vec{j}\cdot\vec{k}=\vec{k}\cdot\vec{i}=0\\ \vec{i}\cdot\vec{i} & =\vec{j}\cdot\vec{j}=\vec{k}\cdot\vec{k}=1 \end{align*}\]
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 \(\left(\vec{j}+2\vec{k}\right)\)
\(\text{(b)$\,$ }\)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 \(\left(\vec{b}-\vec{c}\right)\)?
\(\text{(a)$\,\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}=8\quad\text{$(b)$ $\vec{a}$ is perpendicular to $\left(\vec{b}-\vec{c}\right).$ }$ }\)
