Dot Product and Angle

Given two vectors \(\vec {u}\) and \(\vec {v}\), let \(\theta \) be the angle between them such that \(0\leq \theta \leq \pi \). We will refer to \(\theta \) as the included angle.

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The following theorem establishes a relationship between the dot product and the included angle.

Consider the triangle formed by \(\vec {u}\), \(\vec {v}\) and \(\vec {u}-\vec {v}\).

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By the Law of Cosines we have:

\begin{align*} \norm {\vec {u}-\vec {v}}^2=\norm {\vec {u}}^2+\norm {\vec {v}}^2-2\norm {\vec {u}}\norm {\vec {v}}\cos \theta \end{align*}

By Theorem th:dotproductpropertiesitem:norm of Dot Product and its Properties

\begin{align*} (\vec {u}-\vec {v})\dotp (\vec {u}-\vec {v})=&\vec {u}\dotp \vec {u}+\vec {v}\dotp \vec {v}-2\norm {\vec {u}}\norm {\vec {v}}\cos \theta \end{align*}

By Theorem th:dotproductpropertiesitem:distributive-again of Dot Product and its Properties

\begin{align*} (\vec {u}-\vec {v})\dotp \vec {u}-(\vec {u}-\vec {v})\dotp \vec {v}=&\vec {u}\dotp \vec {u}+\vec {v}\dotp \vec {v}-2\norm {\vec {u}}\norm {\vec {v}}\cos \theta \end{align*}

By Theorem th:dotproductpropertiesitem:distributive of Dot Product and its Properties

\begin{align*} \vec {u}\dotp \vec {u}-\vec {v}\dotp \vec {u}-\vec {u}\dotp \vec {v}+\vec {v}\dotp \vec {v}=&\vec {u}\dotp \vec {u}+\vec {v}\dotp \vec {v}-2\norm {\vec {u}}\norm {\vec {v}}\cos \theta \end{align*}

By Theorem th:dotproductpropertiesitem:commutative of Dot Product and its Properties

\begin{align*} -2(\vec {u}\dotp \vec {v})=&-2\norm {\vec {u}}\norm {\vec {v}}\cos \theta \\ \vec {u}\dotp \vec {v}=&\norm {\vec {u}}\norm {\vec {v}}\cos \theta \end{align*}

Orthogonal Vectors

We can use Theorem 1 to show that two non-zero orthogonal vectors of \(\RR ^n\) are simply perpendicular vectors (the included angle is \(90^{\circ }\)). To see this, suppose that \(\vec {u}\dotp \vec {v}=0\) for nonzero vectors \(\vec {u},\vec {v}\). Then from Theorem 1 we have

\[0=\norm {\vec {u}}\norm {\vec {v}}\cos \theta .\]

Since \(\vec {u},\vec {v}\) are nonzero vectors, we have \(0=\cos \theta \), which implies \(\theta =90^{\circ }\). The converse also holds. If \(\theta =90^{\circ }\), then the dot product is clearly 0.

The reason we prefer the term “orthogonal" to “perpendicular" in this course is because \(\RR ^n\) is only one example of a vector space, and the dot product is only one example of a more general product, called an inner product. For vectors in \(\RR ^n\) a zero dot product happens to coincide with the geometric idea of perpendicularity, but there are many vector spaces that do not possess the visual geometry of \(\RR ^n\). (Later in the text, you will encounter vector spaces whose vectors are polynomial functions!) In these more abstract settings, a zero inner product still signals a special relationship between vectors. The term orthogonal captures this relationship.

Practice Problems

Find the degree measure of the included angle, \(\theta \) for each pair of vectors. Round your answers to the nearest tenth.
1.
\(\begin{bmatrix}1\\2\end{bmatrix}\) and \(\begin{bmatrix}-3\\-1\end{bmatrix}\).

Answer: \(\theta =\answer {135}^\circ \)

2.
\(\begin{bmatrix}-1\\2\\4\end{bmatrix}\) and \(\begin{bmatrix}-2\\1\\-1\end{bmatrix}\)

Answer: \(\theta =\answer {90}^\circ \)

3.
\(\begin{bmatrix}0\\-3\\1\end{bmatrix}\) and \(\begin{bmatrix}-5\\-2\\4\end{bmatrix}\)

Answer: \(\theta =\answer [tolerance=0.1]{61.9}^\circ \)

What does the sign of the dot product tell us about the included angle?
Find all values of \(a\) so that \(\begin{bmatrix}a^2\\2a\\1\end{bmatrix}\) is orthogonal to \(\begin{bmatrix}1\\2\\3\end{bmatrix}\). List your answers in increasing order.

Answer: \(\answer {-3}, \answer {-1}\).

Find the value of \(x\) for which the vector \(\begin{bmatrix}x\\-4\end{bmatrix}\) is parallel to the vector \(\begin{bmatrix}3\\2\end{bmatrix}\). What is the measure of the included angle, \(\theta \)? Find the measure of the included angle using Theorem 3. Do the two results agree?

Answer:

\[x=\answer {-6}\]
\[\theta =\answer {180}^\circ \]
Prove that if \(\vec {u}\) is a unit vector, then \(\vec {u}\dotp \vec {u}=1\).
Prove that if \(\vec {u}_1\) and \(\vec {u}_2\) are unit vectors, then \(-1\leq \vec {u}_1\dotp \vec {u}_2\leq 1\). In what cases are the extreme values of 1 and \(-1\) attained?
Imagine a clock with hands represented by vectors \(\vec {m}\) and \(\vec {h}\), as shown below. At what whole hour will \(\vec {m}\dotp \vec {h}\) attain its maximum value? At what whole hour will \(\vec {m}\dotp \vec {h}\) be as small as possible?

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Answer:

\[\vec {m}\dotp \vec {h}\text { is greatest at }\answer {12}:00 \text { o'clock}\]
\[\vec {m}\dotp \vec {h}\text { is smallest at }\answer {6}:00 \text { o'clock}\]

Photo Credits

The following images are courtesy of Wikimedia Commons

Hannes Grobe, Wall clock manufactured by Telefonbau & Normalzeit. CC-BY 3.0