Extra Topic: Dot Product and Angle

Section 1.4 Extra Topic: Dot Product and Angle

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

$Angle between vectors.$

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

Theorem 1.4.1.

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n\text{,}$ and let $\theta$ be the included angle. Then

\begin{equation} \mathbf{u}\cdot\mathbf{v}=\norm{\mathbf{u}}\norm{\mathbf{v}}\cos \theta\tag{1.4.1} \end{equation}

Proof.

Consider the triangle formed by $\mathbf{u}\text{,}$ $\mathbf{v}$ and $\mathbf{u}-\mathbf{v}\text{.}$

By the Law of Cosines we have:

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

By Theorem 1.2.7 Item 6,

\begin{equation*} (\mathbf{u}-\mathbf{v})\cdot (\mathbf{u}-\mathbf{v})=\mathbf{u}\cdot \mathbf{u}+\mathbf{v}\cdot \mathbf{v}-2\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta \end{equation*}

By Theorem 1.2.7 Item 3,

\begin{equation*} (\mathbf{u}-\mathbf{v})\cdot \mathbf{u}-(\mathbf{u}-\mathbf{v})\cdot \mathbf{v}=\mathbf{u}\cdot \mathbf{u}+\mathbf{v}\cdot \mathbf{v}-2\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta \end{equation*}

By [cross-reference to target(s) "th-dotproductpropert, ies" missing or not unique] Item 2,

\begin{equation*} \mathbf{u}\cdot \mathbf{u}-\mathbf{v}\cdot\mathbf{u}-\mathbf{u}\cdot\mathbf{v}+\mathbf{v}\cdot \mathbf{v}=\mathbf{u}\cdot \mathbf{u}+\mathbf{v}\cdot \mathbf{v}-2\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta \end{equation*}

By Theorem 1.2.7 Item 1 of [cross-reference to target(s) "Section-Dot-Product-and-its-Properties" missing or not unique],

\begin{align*} -2(\mathbf{u}\cdot \mathbf{v})=\amp-2\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta \\ \mathbf{u}\cdot \mathbf{v}=\amp\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta \end{align*}

Example 1.4.2.

Find the included angle between vectors $\mathbf{u}=\begin{bmatrix}2\\-1\\4\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}1\\2\\-3\end{bmatrix}\text{.}$

By Theorem 1.4.1, $\cos \theta=\frac{\mathbf{u}\cdot \mathbf{v}}{\norm{\mathbf{u}}\norm{\mathbf{v}}}\text{.}$

\begin{equation*} \mathbf{u}\cdot\mathbf{v}=2-2-12=-12 \end{equation*}

\begin{equation*} \norm{\mathbf{u}}=\sqrt{4+1+16}=\sqrt{21} \end{equation*}

\begin{equation*} \norm{\mathbf{v}}=\sqrt{1+4+9}=\sqrt{14} \end{equation*}

\begin{equation*} \cos\theta =\frac{-12}{\sqrt{21}\sqrt{14}} \end{equation*}

\begin{equation*} \theta \approx 134.4^{\circ} \end{equation*}

Subsection 1.4.1 Orthogonal Vectors

Definition 1.4.3.

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n\text{.}$ We say $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if $\mathbf{u}\cdot \mathbf{v}=0\text{.}$

We can use Theorem 1.4.1 to show that two non-zero orthogonal vectors of $\R^n$ are simply perpendicular vectors (the included angle is $90^{\circ}$). To see this, suppose that $\mathbf{u}\cdot\mathbf{v}=0$ for nonzero vectors $\mathbf{u},\mathbf{v}\text{.}$ Then from Theorem 1.4.1 we have

\begin{equation*} 0=\norm{\mathbf{u}}\norm{\mathbf{v}}\cos\theta. \end{equation*}

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

The reason we prefer the term ``orthogonal" to ``perpendicular" in this course is because $\R^n$ 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 $\R^n$ 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 $\R^n\text{.}$ (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.

Remark 1.4.4.

By our definition, the zero vector is orthogonal to any vector. However, we will not use the word perpendicular when the zero vector is involved, as it is not possible to talk about an ``included angle’’.

Exercises 1.4.2 Exercises

Exercise Group.

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 $\begin{bmatrix}-3\\-1\end{bmatrix}\text{.}$

Answer.

$\theta=135^\circ$

2.

$\begin{bmatrix}-1\\2\\4\end{bmatrix}$ and $\begin{bmatrix}-2\\1\\-1\end{bmatrix}$

Answer.

$\theta=90^\circ$

3.

$\begin{bmatrix}0\\-3\\1\end{bmatrix}$ and $\begin{bmatrix}-5\\-2\\4\end{bmatrix}$

Answer.

$\theta= 61.9^\circ$

4.

$\begin{bmatrix}1\\1\\-1\\2\end{bmatrix}$ and $\begin{bmatrix}-2\\0\\-3\\1\end{bmatrix}$

Answer.

$\theta=72.4^\circ$

5.

What does the sign of the dot product tell us about the included angle?

6.

Find all values of $a$ so that $\begin{bmatrix}a^2\\2a\\1\end{bmatrix}$ is orthogonal to $\begin{bmatrix}1\\2\\3\end{bmatrix}\text{.}$

Answer.

$a=-3, -1$

7.

Find the value of $x$ for which the vector $\begin{bmatrix}x\\-4\end{bmatrix}$ is parallel to the vector $\begin{bmatrix}3\\2\end{bmatrix}\text{.}$ What is the measure of the included angle, $\theta\text{?}$ Find the measure of the included angle using Example 1.4.2. Do the two results agree?

Answer.

$x=-6$ and $\theta=180^\circ$

8.

Prove that if $\mathbf{u}$ is a unit vector, then $\mathbf{u}\cdot \mathbf{u}=1\text{.}$

9.

Prove that if $\mathbf{u}_1$ and $\mathbf{u}_2$ are unit vectors, then $-1\leq\mathbf{u}_1\cdot \mathbf{u}_2\leq 1\text{.}$ In what cases are the extreme values of 1 and $-1$ attained?

10.

Imagine a clock with hands represented by vectors $\mathbf{m}$ and $\mathbf{h}\text{,}$ as shown below. At what whole hour will $\mathbf{m}\cdot\mathbf{h}$ attain its maximum value? At what whole hour will $\mathbf{m}\cdot\mathbf{h}$ be as small as possible?

Image of a wall clock. — Figure 1.4.5. Image courtesy of Wikipedia Commons
¹
`commons.wikimedia.org/wiki/Main_Page`
.

Answer.

\begin{equation*} \mathbf{m}\cdot\mathbf{h}\text{ is greatest at }12:00 \text{ o'clock} \end{equation*}

\begin{equation*} \mathbf{m}\cdot\mathbf{h}\text{ is smallest at }6:00 \text{ o'clock} \end{equation*}

11.

Let $O$ be a circle of radius $r\text{.}$ Suppose $A$ and $B$ are the endpoints of a diameter of $O\text{,}$ and $C$ is a point on $O$ distinct from $A$ and $B\text{.}$ Show that vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$ are orthogonal.

$Circumscribed circle of a triangle.$

Hint.

Assign coordinates to points $A\text{,}$ $B$ and $C\text{,}$ express vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$ in component form, then find the dot product of $\overrightarrow{AC}$ and $\overrightarrow{BC}\text{.}$

12.

A rhombus is a quadrilateral with four congruent sides. Use vectors to prove that a parallelogram is a rhombus if and only if its diagonals are perpendicular.

Hint.

See section on vector subtraction in [cross-reference to target(s) "Section-Vector-Arithmetic" missing or not unique].

13.

The points $A(2,1,4),B(4,2,-1)\text{,}$ and $C(6,8,7)$ form a triangle in $\R^3\text{.}$ Is it a right triangle?

Hint.

Express each side of the triangle as a vector and use what you have learned in this section.

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