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.
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.
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.
- \(\vec{u} \cdot \vec{v}=\vec{v} \cdot \vec{u}\)the dot product is commutative
- \(\vec{u} \cdot(\vec{v}+\vec{w})=\vec{u} \cdot \vec{v}+\vec{u} \cdot \vec{w}\) the dot product distributes over vector addition
- \(\vec{u} \cdot \overrightarrow{0}=0\),the dot product with the zero vector \(\overrightarrow{0}\), is the scalar 0.
- \(\vec{u} \cdot \vec{u}=\|\vec{u}\|^2\)
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.
\[\begin{equation}
\|\vec{v}\|=\sqrt{v_x^2+v_y^2}
\end{equation} \nonumber \]
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.
If \(\theta\) is the smallest nonnegative angle between two non-zero vectors \(\vec{u}\) and \(\vec{v}\) then
\[\begin{equation}
\cos \theta=\frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|} \; \text { or } \; \theta=\cos ^{-1} \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \cdot\|\vec{v}\|}
\end{equation} \nonumber \]
\[\begin{equation}
\text { where } \; 0 \leq \theta \leq 2 \pi \; \text { and } \; \|\vec{u}\|=\sqrt{u_x^2+u_y{ }^2} \; \text { and } \; \|\vec{v}\|=\sqrt{v_x^2+v_y{ }^2}
\end{equation} \nonumber \]
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
Find the dot product of the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
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\(\overrightarrow{u}\bullet \overrightarrow{v}=-13\)
Find the dot product of the vectors \(\vec{u}=\langle-4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
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\(\overrightarrow{u}\bullet \overrightarrow{v}=0\)
Find the length of the vector \(\vec{u}=\langle 4,-7\rangle \).
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\(\sqrt{65}\)
Find the length of the vectors \(\vec{v}=\langle 0,5\rangle \).
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\(5\)
Find the angle between the vectors \(\vec{u}=\langle-2,3\rangle \) and \(\vec{v}=\langle 5,-1\rangle \).
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\(135{}^\circ\)
Find the angle between the vectors \(\vec{u}=\langle -4,6\rangle \) and \(\vec{v}=\langle 3,2\rangle \).
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\(90{}^\circ\)