The first type of vector multiplication is called the dot product. The dot product of vectors \(\vec A\) and \(\vec B\) is the sum of the products of each pair of components. That is,

$$\vec A\cdot\vec B=A_xB_x+A_yB_y+A_zB_z$$

Notice that the result of the dot product is a
$$\vec A\cdot\vec B=\abs{\vec A}\abs{\vec B}\cos\theta$$

where - If two vectors point in the same direction (
*θ*= 0°), then their dot product is just the product of their magnitudes. In particular$$\vec A\cdot\vec A=\abs{\vec A}^2$$ - If two vectors point in the same-ish direction (that is, if the angle between them is less than 90°), then their dot product is
*positive*because the cosine of an acute angle is positive. - The dot product of two vectors at right angles to each other is zero. For example, \(\hat x\cdot\hat y=0\).
- If the angle between two vectors is
*obtuse*(i.e. greater than 90°), so that they point in opposite-ish directions, then their dot product is*negative*.

Find the dot product of \((3\hat x+\hat y)\) and \((-\hat x+4\hat y)\). Do the two vectors point in the same-ish or opposite-ish directions?

Multiply the like components and add them together:
$$(3)(-1)+(1)(4)=-3+4=\fbox{1}$$
The dot product is positive, so the vectors point in the same-ish direction (that is, the angle between them is acute.)

Find the angle between the vectors \(\hat y\) and \(3\hat x+4\hat y\).

The dot product of the two vectors is
$$(3)(0)+(1)(4)=4$$
But the dot product is also the product of the magnitude of the two vectors times the cosine of the angle between them. Thus
$$4=|\hat y||3\hat x+4\hat y|\cos\theta$$
The magnitude of \(\hat y\) is 1, and the magnitude of \(3\hat x+4\hat y\) is \(\sqrt{3^2+4^2}=5\), and so
$$4=5\cos\theta \implies \cos\theta={4\over5}$$
$$\implies \theta=\cos^{-1}0.8=\fbox{$36.9^\circ$}$$

Vectors \(3\hat x+4\hat y\) and \(4\hat x+a\hat y\) are perpendicular; find \(a\).

If they are perpendicular, their dot product must be zero:
$$\begin{eqnarray}
0&=&(3\hat x+4\hat y)\cdot(4\hat x+a\hat y)\cr
&=&12+4a\cr
4a&=&-12\cr
a&=&\fbox{$-3$}\cr
\end{eqnarray}
$$
Therefore \(3\hat x+4\hat y\) and \(4\hat x-3\hat y\) are perpendicular to each other. In general,
if you have a vector \(a\hat x+b\hat y\) and you need another vector which is perpendicular to it, you can use \(b\hat x-a\hat y\).

Find the angle

The vector \(\vec F=\hat x+\hat z\) points along the face diagonal, and the vector \(\vec B=\hat x+\hat y+\hat z\) points along the body diagonal; thus the angle between these two vectors is the angle *θ* we seek. From the definition of the dot product
$$\vec F\cdot\vec B=|\vec F||\vec B|\cos\theta \implies \theta=\cos^{-1}\left(\frac{\vec F\cdot\vec B}{|\vec F||\vec B|}\right)$$
The magnitude of
\(\vec F\) is \(|{\vec F}|=\sqrt{1^2+0^2+1^2}=\sqrt{2}\); the magnitude of \(\vec B\) is \(\sqrt{3}\). The dot product between the two is
\((\hat x+\hat z)\cdot(\hat x+\hat y+\hat z)=1(1)+0(1)+1(1)=2\).
Thus
$$\theta=\cos^{-1}\left(\frac{2}{\sqrt{2}\sqrt{3}}\right)=\cos^{-1}0.8165=\fbox{$35.2^\circ$}$$

Interactive 1.4.1