2.4: The Dot Product of Two Vectors, the Length of a Vector, and the Angle Between Two Vectors

The Dot Product of Two Vectors

The length of a vector or the angle between two vectors \(\vec{u}\boldsymbol{=}\left\langle u_x,\left.u_y\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) can be found using the dot product.

Definition: Dot Product of Vectors

The dot product of vectors \(\vec{u}=\left\langle u_x, u_y\right\rangle\) and \(\vec{v}=\left\langle v_x, v_y\right\rangle\) is a scalar (real number) and is defined to be \[\vec{u} \cdot \vec{v}=u_x v_x+u_y v_y \nonumber \]

Since \(u_x,\ u_y,\ v_x\) and \(v_y\) are real numbers, you can see that the dot product is itself a real number and not a vector.

Example \(\PageIndex{1}\)

To compute the dot product of the vectors \(\vec{u}\boldsymbol{=}\left\langle 5,\left.2\right\rangle \right.\) and \(\vec{v}\boldsymbol{=}\left\langle 3,\left.4\right\rangle \right.\)

Solution

We compute: \[\overrightarrow{u}\bullet \overrightarrow{v}\boldsymbol{=}5\bullet 3\mathrm{+\ }2\bullet 4=15+8=23 \nonumber \]

Since the dot product is a scalar, it follows the properties of real numbers.

Properties of the Dot Product

1. \(\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}\)the dot product is commutative
2. \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w}\) the dot product distributes over vector addition
3. \(\vec{u} \cdot \overrightarrow{0}=0\),the dot product with the zero vector \(\overrightarrow{0}\), is the scalar 0.
4. \(\vec{u} \cdot \vec{u}=\|\vec{u}\|^2\)

Example \(\PageIndex{2}\)

Compute the dot product \(\overrightarrow{u}\bullet \left(\overrightarrow{v}+\overrightarrow{w}\right)=\overrightarrow{u}\bullet \overrightarrow{v}+\overrightarrow{u}\bullet \overrightarrow{w}\), where \(\overrightarrow{u}=\left\langle 5,\left.-2\right\rangle \right.\), \(\overrightarrow{v}=\left\langle 6,\left.4\right\rangle \right.\), and \(\overrightarrow{w}=\left\langle -3,\left.7\right\rangle \right.\).

Solution

\[\begin{align} \vec{u} \cdot(\vec{v}+\vec{w})&=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w} \nonumber\\
\vec{u} \cdot(\vec{v}+\vec{w})&=\langle 5,-2\rangle \cdot\langle 6,4\rangle+\langle 5,-2\rangle \cdot\langle-3,7\rangle \nonumber \\
&=(5 \cdot 6+(-2) \cdot 4)+(5 \cdot(-3)+(-2) \cdot 7) \nonumber \\
&=30-8-15-14 \nonumber \\
&=-7 \nonumber
\end{align} \nonumber \]

The Length of a Vector

The length (magnitude) of a vector you know is given by \(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\). The length can also be found using the dot product. If we dot a vector \(\overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\) with itself, we get

\[\overrightarrow{v}\bullet \overrightarrow{v}=\left\langle v_x,\left.v_y\right\rangle \right.\bullet \left\langle v_x,\left.v_y\right\rangle \right. \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}=v_x\bullet v_x+v_y\bullet v_y \nonumber \] \[\overrightarrow{v}\bullet \overrightarrow{v}= {v_x}^2+\ {v_y}^2 \nonumber \]

By Vector Property 4, \(\overrightarrow{v}\bullet \overrightarrow{v}=\ {\left\|\overrightarrow{v}\right\|}^2\). This gives \({\left\|\overrightarrow{v}\right\|}^2={v_x}^2+{v_y}^2\).

Taking the square root of each side produces

\[\sqrt{{\|\overrightarrow{v}\|}^2}=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]

\[\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2} \nonumber \]

Which is the length of the vector \(\overrightarrow{v}\).

The dot product of a vector \(\vec{v}=\left\langle v_x, v_y\right\rangle\) with itself gives the length of the vector.

\[\begin{equation}
\|\vec{v}\|=\sqrt{v_x^2+v_y^2}
\end{equation} \nonumber \]

Example \(\PageIndex{3}\)

Use the dot product to find the length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\).

Solution

In this case, \(v_x=2\) and \(v_y=6.\)

Using \(\|\overrightarrow{v}\|=\sqrt{{v_x}^2+{v_y}^2}\), we get \[\|\overrightarrow{v}\|=\sqrt{2^2+6^2} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{40} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4\bullet 10} \nonumber \] \[\|\overrightarrow{v}\|=\sqrt{4}\bullet \sqrt{10} \nonumber \] \[\|\overrightarrow{v}\|=2\sqrt{10} \nonumber \]

The length of the vector \(\overrightarrow{v}=\left[ \begin{array}{c} 2 \\ 6 \end{array} \right]\) is \(2\sqrt{10}\) units.

The Angle between two Vectors

The dot product and elementary trigonometry can be used to find the angle \(\theta\) between two vectors.

Definition: Angle between two Vectors

If \(\theta\) is the smallest nonnegative angle between two non-zero vectors \(\vec{u}\) and \(\vec{v}\) then

\[\begin{equation}
\cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \; \text { or } \; \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|}
\end{equation} \nonumber \]

\[\begin{equation}
\text { where } \; 0 \leq \theta \leq 2 \pi \; \text { and } \; \|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2} \; \text { and } \; \|\vec{v}\|=\sqrt{v_x^2+v_y{ }^2}
\end{equation} \nonumber \]

Example \(\PageIndex{4}\)

Find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\).

Solution

Using \(\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\overrightarrow{u}\bullet \overrightarrow{v}}{\|\overrightarrow{u}\|\bullet \|\overrightarrow{v}\|}\), we get

\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{\left\langle 5,\left.-3\right\rangle \right.\bullet \left\langle 2,\left.4\right\rangle \right.}{\sqrt{5^2+{\left(-3\right)}^2}\bullet \sqrt{2^2+4^2}} \nonumber \]

\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{5\bullet 2+\left(-3\right)\bullet 4}{\sqrt{25+9}\bullet \sqrt{4+16}} \nonumber \]

\[\theta =\ {\mathrm{cos}}^{\mathrm{-}\mathrm{1}}\frac{-2}{\sqrt{34}\bullet \sqrt{20}} \nonumber \]

\[\theta =94.4 \nonumber \]

We conclude that the angle between these two vectors is close to 94.4\(\mathrm{{}^\circ}\).

Using Technology

We can use technology to find the angle \(\theta\) between two vectors.

Go to www.wolframalpha.com.

To find the angle between the vectors \(\overrightarrow{u}=\left\langle 5,\left.-3\right\rangle \right.\) and \(\overrightarrow{v}=\left\langle 2,\left.4\right\rangle \right.\), enter angle between the vectors \(\mathrm{<}\)5, -3\(\mathrm{>}\) and \(\mathrm{<}\)2, 4\(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\theta =94.4\), rounded to one decimal place.

Try These

Exercise \(\PageIndex{1}\)

Find the dot product of the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).

Answer

\(\overrightarrow{u}\bullet \overrightarrow{v}=-13\)

Exercise \(\PageIndex{2}\)

Find the dot product of the vectors \(\vec{u}=\langle-4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).

Answer

\(\overrightarrow{u}\bullet \overrightarrow{v}=0\)

Exercise \(\PageIndex{3}\)

Find the length of the vector \(\vec{u}=\langle 4,-7\rangle \).

Answer

\(\sqrt{65}\)

Exercise \(\PageIndex{4}\)

Find the length of the vectors \(\vec{v}=\langle 0,5\rangle \).

Answer

\(5\)

Exercise \(\PageIndex{5}\)

Find the angle between the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).

Answer

\(135{}^\circ\)

Exercise \(\PageIndex{6}\)

Find the angle between the vectors \(\vec{u}=\langle -4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).

Answer

\(90{}^\circ\)