5. Vectors

d. Dot Product

4. Angles

The reason the dot product is useful is that the geometric relationship \[ \vec u\cdot\vec v=|\vec u|\,|\vec v|\cos\theta \] allows us to find the angle between two vectors.

The angle, \(\theta\), between the vectors \(\vec u\) and \(\vec v\) satisfies: \[ \cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u|\,|\vec v|} \] Memorize this!

Find the angle between the two vectors \(\langle \sqrt{3},1\rangle\) and \(\langle 1,\sqrt{3}\rangle\).

From the formula, we have \[ \cos\theta =\dfrac{\langle \sqrt{3},1\rangle\cdot\langle 1,\sqrt{3}\rangle} {|\langle \sqrt{3},1\rangle|\,|\langle 1,\sqrt{3}\rangle|} =\dfrac{2\sqrt{3}}{2\cdot2}=\dfrac{\sqrt{3}}{2} \] Thus, \(\theta=30^\circ=\dfrac{\pi}{6}\,\text{rad}\).

Find the angle between the two vectors \(\vec a=\left\langle 2,2\right\rangle\) and \(\vec b=\left\langle 0,3\right\rangle\).

\(\theta=45^\circ=\dfrac{\pi}{4}\,\text{rad}\)

We compute: \[\begin{aligned} |\vec a|=\sqrt{4+4}=2\sqrt{2}& \qquad |\vec b|=\sqrt{0+9}=3 \\ \text{and} \quad \vec a\cdot\vec b&=0+6=6 \end{aligned}\] So: \[ \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|} =\dfrac{6}{2\sqrt{2}\cdot3}=\dfrac{1}{\sqrt{2}} \] Thus, \(\theta=45^\circ=\dfrac{\pi}{4}\,\text{rad}\).

Find the angle between the two vectors \(\vec p=\langle -\sqrt{3},1\rangle\) and \(\vec q=\langle 1,\sqrt{3}\rangle\).

\(\theta=90^\circ=\dfrac{\pi}{2}\,\text{rad}\) The vectors are perpendicular.

We compute: \[\begin{aligned} |\vec p|=\sqrt{3+1}=2& \qquad |\vec q|=\sqrt{1+3}=2 \\ \text{and} \quad \vec p\cdot\vec q&=-\sqrt{3}+\sqrt{3}=0 \end{aligned}\] So: \[ \cos\theta=\dfrac{\vec p\cdot\vec q}{|\vec p|\,|\vec q|}=0 \] Thus, \(\theta=90^\circ=\dfrac{\pi}{2}\,\text{rad}\). The vectors are perpendicular.

You can also practice computing an angle of a triangle using vectors and dot products in 3D by using the following Maplet (requires Maple on the computer where this is executed):

Angles of Triangles in 3D Space Rate It

5. Vectors

d. Dot Product

4. Angles

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