Scalar product and angles

Consider two (non-zero) vectors $\vec{a}$ and $\vec{b}$ and represent them in space as arrows with overlapping tails.

We want to find the (smaller) angle between the two vectors (which will always be between $0^\circ$ and $180^\circ$ ). It turns out that the angle is closely related to the scalar product between $\vec a$ and $\vec b$ :

\boxed{\cos(\alpha)=\frac{\vec{a}\bullet \vec{b}}{\vert\vec{a}\vert\cdot \vert\vec{b}\vert} }

Once we have calculated the value for $\cos(\alpha)$ , we can calculate $\alpha$ using $\cos^{-1}$ .

Example 1

Consider the vectors $\vec{a} = \left(\begin{array}{r} 3\\ 5\\ 2 \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} -1\\ 3\\ 8 \end{array}\right)$ . It is

\cos(\alpha)=\frac{\vec{a}\bullet \vec{b}}{\vert\vec{a}\vert\cdot \vert\vec{b}\vert} = \frac{3\cdot (-1)+5\cdot 3+2\cdot 8}{\sqrt{3^2+5^2+2^2}\cdot \sqrt{(-1)^2+3^2+8^2}} = \frac{28}{\sqrt{38}\cdot \sqrt{74}} = 0.528

Thus, the smaller angle between $\vec a$ and $\vec b$ is $\alpha = \cos^{-1}(0.528)=58.12^\circ$

Exercise 1

Consider the vectors $\vec{a}=\left(\begin{array}{r} 0\\ 1\\ 4 \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} 0\\ 3\\ 1 \end{array}\right)$ .

Draw the arrows and measure the angle between the two arrows.
Calculate the angle and compare with the measurement.

Solution