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# 5.4: Dot Products

Difficulty Level: At Grade Created by: CK-12
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Kelly is learning about flying airplanes so she can apply to take her pilot's exam. Right now she is studying the effect of wind on a flying plane, and she is supposed to describe the added "push" effect a NNE wind of 50mph would have on a plane travelling at 30mph due NE relative the ground.

How can she describe the combination of forces in the NE direction?

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### Guidance

In common terms, the dot product of two vectors is a number that describes the amount of force that two different vectors both contribute in the direction. This value results from the multiplication of the appropriate parts of each vector, and is therefore sometimes referred to as the scalar product of the two vectors.

A dot product is a scalar quantity which varies as the angle between the two vectors changes. The angle between the vectors affects the dot product because the portion of the total force of a vector dedicated to a particular direction goes up or down if the entire vector is pointed toward or away from that direction.

The maximum value for the dot product occurs when the two vectors are parallel to one another (all 'force' from both vectors is in the same direction), but when the two vectors are perpendicular to one another, the value of the dot product is equal to 0 (one vector has zero force aligned in the direction of the other, and any value multiplied by zero is zero).

There are two ways to calculate the dot product. One way is to multiply the individual components. Each component of vector \begin{align*}\overrightarrow{A}\end{align*} is multiplied by the component of vector\begin{align*}\overrightarrow{B}\end{align*} which is in the same direction. Then we add the results.

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = {A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z} + ...\end{align*}

Another way to describe the process is to say that the dot product is the multiplication of one vector by the component of a second vector which is parallel to the first vector. In the diagram below are two vectors, A and B. A perpendicular line has been drawn radially outward from B towards A to create a right triangle with A as the hypotenuse.

The component of\begin{align*}\overrightarrow{A}\end{align*} which is parallel to \begin{align*}\overrightarrow{B}\end{align*} is given by A cos θ so the second way to compute the dot product is \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{A}(\overrightarrow{B}\ \mbox{cos}\ \theta) = | \overrightarrow{A} ||\overrightarrow{B}|\ \mbox{cos}\ \theta\end{align*}

Likewise, the component of \begin{align*}\overrightarrow{B}\end{align*} which is parallel to \begin{align*}\overrightarrow{A}\end{align*} is given by B cos θ , so the dot product \begin{align*}\overrightarrow{B} \times \overrightarrow{A} = |\overrightarrow{B}|(|\overrightarrow{A}|\ \mbox{cos}\ \theta) = |\overrightarrow{A}||\overrightarrow{B}|\ \mbox{cos}\ \theta\end{align*}

No matter which of the two vectors we "project" onto the other, the value of the dot product is maximized when the two vectors are parallel and zero when the two vectors are perpendicular to one another. When a vector is dotted with itself, the result is the square of the vector's magnitude since, by definition, a vector has the same direction as an equal vector.

\begin{align*}\overrightarrow{A} \times \overrightarrow{A} = |\overrightarrow{A}|\ (|\overrightarrow{A}|\ \mbox{cos}\ \theta) = |\overrightarrow{A}||\overrightarrow{A}|\ \mbox{cos}\ 0 = |\overrightarrow{A}|^2\end{align*}

Naturally the dot product of any vector with a zero vector is zero since the magnitude of the zero vector is 0.

#### Example A

(a) Calculate the dot product of the two vectors shown below. (b) Determine the angle between the two vectors.\begin{align*}\overrightarrow{A} = \left \langle {1},{3} \right \rangle \end{align*} and \begin{align*}\overrightarrow{B} = \left \langle {3},{2} \right \rangle\end{align*}.

Solution

First we will use the components of the two vectors to determine the dot product.

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = {A_x}{B_x} + {A_y}{B_y} = (1 \cdot 3) + (3 \cdot 2) = 3 + 6 = 9\end{align*}

Now that we know the dot product, the alternative definition of the dot product, \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = |A||B|\ \mbox{cos}\ \theta\end{align*} to find θ, the angle between the vectors. First find the magnitudes of the two vectors:

\begin{align*}|A| = \sqrt{{A_x}^2 + {A_y}^2} = \sqrt{{1^2}+{3^2}}=\sqrt{{1}+{9}}=\sqrt{10}\end{align*}

\begin{align*}|B| = \sqrt{{B_x}^2 + {B_y}^2} = \sqrt{{3^2}+{2^2}}=\sqrt{{9}+{4}}=\sqrt{13}\end{align*}

Then use these magnitudes with the cosine version of the dot product to find θ.

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = |A||B|\ \mbox{cos}\ \theta\end{align*}

\begin{align*}9 = \sqrt{10}\sqrt{13} \ \mbox{cos}\ \theta\end{align*}

\begin{align*}\mbox{cos}\ \theta = \frac{9} {\sqrt{130}} = \frac{9}{11.4} = 0.78935\end{align*}

\begin{align*}\theta = 37.9^\circ\end{align*}

#### Example B

Calculate the dot product of the two vectors shown below. Then determine the angle between the two vectors.

Solution

Use the components of the two vectors to determine the dot product.

Here \begin{align*}\overrightarrow{A} = \left \langle {2,3} \right \rangle\end{align*} and \begin{align*}\overrightarrow{B} = \left \langle {3,-1} \right \rangle\end{align*}.

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = {A_x}{B_x}+{A_y}{B_y}=(2 \cdot 3)+(3 \cdot -1)=6+(-3)=3\end{align*}

Now to find the angle between the vectors, first find the magnitudes of the two vectors:

\begin{align*}|A|=\sqrt{{A_x}^2+{A_y}^2}=\sqrt{{2^2}+{3^2}}=\sqrt{{4}+{9}}=\sqrt{13}\end{align*}

\begin{align*}|B|=\sqrt{{B_x}^2+{B_y}^2}=\sqrt{{3^2}+{(-1)^2}}=\sqrt{{9}+{1}}=\sqrt{10}\end{align*}

Then use these magnitudes with the cosine version of the dot product to find θ.

\begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|A||B|\ \mbox{cos}\ \theta\end{align*}

\begin{align*}3=\sqrt{10}\sqrt{13}\ \mbox{cos}\ \theta\end{align*}

\begin{align*}\mbox{cos}\ \theta = \frac{3} {\sqrt{130}} = \frac{3}{11.4}=0.26312\end{align*}

\begin{align*}\theta = 74.7^\circ\end{align*}

#### Example C

The diagram below shows both vectors AB and D together on the same grid. Determine the scalar projection of vector AB onto the direction of vector D.

Solution

To find the scalar projection onto the direction of another vector we need to know the unit vector in the direction of vector D.

First, the components of \begin{align*}\overrightarrow{D}\end{align*} are

\begin{align*}\overrightarrow{D}=\left \langle 3,-1.25 \right \rangle\end{align*}

Now the magnitude of \begin{align*}\overrightarrow{D}\end{align*} is

\begin{align*}|D|=\sqrt{(D_x)^2+(D_y)^2}=\sqrt{3^2+(-1.25)^2}=\sqrt{9+1.5625}=\sqrt{10.5625}\end{align*} \begin{align*}=3.25\end{align*}.

Finally, the direction vector of \begin{align*}\overrightarrow{D}\end{align*} is

\begin{align*}\overrightarrow{D}=\frac{\overrightarrow{D}}{|D|}=\frac{3\hat{x}+(-1.25)\hat{y}}{3.25}=\frac{3}{3.25}\ \hat{x}+\frac{-1.25}{3.25}\ \hat{y}\end{align*}.

\begin{align*}\overrightarrow{D}=\left \langle 0.923,-0.385 \right \rangle\end{align*}

Now we can use the dot-product to calculate the scalar projection of AB onto the direction of vector D.

\begin{align*}\overrightarrow{AB}\ \times \overrightarrow{D}=(3 \cdot 0.923)+(4 \cdot -0.385)+(0 \cdot 0)=2.769+(-1.54)=1.23\end{align*}

Concept question wrap-up

"Kelly is learning about flying airplanes so she can apply to take her pilot's exam. Right now she is studying the effect of wind on a flying plane, and she is supposed to describe the added "push" effect a NNE wind of 50mph would have on a plane travelling at 30mph due NE relative the ground. How can she describe the combination of forces in the NE direction?"

Kelly could calculate the dot product of the two vectors and use the result to describe the total "push" in the NE direction.

### Vocabulary

A scalar product is the same as a dot product, and is the product of the component of one vector which is parallel to a 2nd vector and the magnitude of the 2nd vector.

### Guided Practice

Questions

1) Determine the dot product of the two vectors \begin{align*}\overrightarrow{f}=\left \langle 3,13,11 \right \rangle\end{align*} and \begin{align*}\overrightarrow{g}=\left \langle 9,6,15 \right \rangle\end{align*}

2) Determine the dot product of the two vectors shown in the diagram below. \begin{align*}\overrightarrow{A}=7cm\end{align*} @ \begin{align*}0^\circ\end{align*} and \begin{align*}\overrightarrow{B}=4cm\end{align*} @ \begin{align*}-22^\circ\end{align*}

3) Determine the dot product of two vectors \begin{align*}\overrightarrow{E}=\left \langle 14,8.5,21 \right \rangle\end{align*} and \begin{align*}\overrightarrow{G}=\left \langle15,12.4,-3.7 \right \rangle\end{align*} . Then determine the angle between the two vectors.

4) Determine the dot product of the following two vectors.

5) Determine the vector projection of vector \begin{align*}\overrightarrow{MN}\end{align*} onto the vector \begin{align*}\overrightarrow{KL}\end{align*} .

6) Determine the dot product of the two vectors \begin{align*}\overrightarrow{w}=\left \langle 85,89,91 \right \rangle\end{align*} and \begin{align*}\overrightarrow{h}=\left \langle 67,70,88 \right \rangle\end{align*} , then determine the angle between the two vectors.

Solutions

1) The component form of the dot product is given by \begin{align*}\overrightarrow{f} \times \overrightarrow{g}=f_x g_x+f_y g_y+f_z g_z\end{align*} . In this case, \begin{align*}\overrightarrow{f} \times \overrightarrow{g}=(3 \cdot 9)+(13 \cdot 6)+(11 \cdot 15)=27+78+165=270\end{align*}

2) The angle form of the dot product is given by \begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|\ \mbox{cos}\ \theta\end{align*}. In this case, \begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|\ \mbox{cos}\ \theta =(7)(4)\ \mbox{cos}\ 22 =28\ \mbox{cos}\ 22=25.96 \ cm^2\end{align*}

3) The dot product of two vectors is defined in two ways: \begin{align*}\overrightarrow{A} \times \overrightarrow{B}=A_x B_x+A_y B_y+A_z B_z+...\end{align*} and \begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|A||B|\ \mbox{cos}\ \theta\end{align*} . We will use the first to calculate the dot product and then we will use that result together with the second definition to determine the angle between the two vectors.

\begin{align*}\overrightarrow{E} \times \overrightarrow{G}=E_x G_x + E_y G_y + E_z G_z\end{align*}

\begin{align*}\overrightarrow{E} \times \overrightarrow{G}=(14)(15)+(8.5)(12.4)+(21)(-3.7)=210+105.4-77.7=237.7\end{align*}

To find the angle between the two vectors, we need to know not only the dot product of the two vectors, but also the length of each individual vector.

\begin{align*}|\overrightarrow{E}| = \sqrt{E^2_x + E^2_y + E^2_z} = \sqrt{(14)^2 + (8.5)^2 + (21)^2} = \sqrt{196 + 72.25 + 441}\end{align*} \begin{align*}= \sqrt{709.25} = 26.63\end{align*}

\begin{align*}|\overrightarrow{G}| = \sqrt{G^2_x + G^2_y + G^2_z} = \sqrt{(15)^2 + (12.4)^2 + (-3.7)^2} = \sqrt{225 + 153.76 + 13.69}\end{align*} \begin{align*}= \sqrt{392.45} = 19.81\end{align*}

Now use the second definition of the dot product to determine the angle

\begin{align*}\overrightarrow{E} \times \overrightarrow{G} = | E | | G |\ cos\ \theta\end{align*}

\begin{align*}\mbox{cos}\ \theta = \frac{\overrightarrow{E} \times \overrightarrow{G}} {| E | | G |} = \frac{237.7} {(26.63)(19.81)} = \frac{237.7} {527.54} = 0.45058\end{align*}

\begin{align*}\theta=\mbox{cos}^{-1} (0.45058)=63.2^\circ\end{align*}

4) The angle form of the dot product is given by \begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|\overrightarrow{A}| |\overrightarrow{B}| \ \mbox{cos}\ \theta\end{align*} . In this case,

\begin{align*}\overrightarrow{A} \times \overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|\ \mbox{cos}\ \theta=(61)(45)\ \mbox{cos}\ 58=2745\ \mbox{cos}\ 58=1455\end{align*}

5) Determine the angle between the vectors \begin{align*}\overrightarrow{MN}\end{align*} and \begin{align*}\overrightarrow{KL}\end{align*} .

We calculated the dot product of \begin{align*}\overrightarrow{MN}\end{align*} and \begin{align*}\overrightarrow{KL}\end{align*} in the previous problem:

\begin{align*}\overrightarrow{MN} \times \overrightarrow{KL}=(MN)_x\ (KL)_x\ + \ (MN)_y\ (KL)_y\ + \ (MN)_z\ (KL)_z\end{align*} \begin{align*}= (2.25)(1.5) + (2)(0) + (0)(0)\end{align*} \begin{align*}= 3.375\end{align*}

We can then use the definition \begin{align*}\overrightarrow{MN} \times \overrightarrow{KL}=|\overrightarrow{MN}| |\overrightarrow{KL}| \ \mbox{cos}\ \theta\end{align*} to determine the angle between the two vectors. But first we need to determine the magnitudes of the two vectors.

\begin{align*}|\overrightarrow{MN}|=\sqrt{(MN)_x^2+(MN)_y^2+(MN)_z^2}=\sqrt{(-2.25)^2+(0)^2+(0)^2}\end{align*} \begin{align*}= 2.25\end{align*}

\begin{align*}|\overrightarrow{KL}|=\sqrt{(KL)_x^2+(KL)_y^2+(KL)_z^2}=\sqrt{(1.5)^2+(2)^2+(0)^2}=\sqrt{2.25+4+0}\end{align*} \begin{align*}= \sqrt{6.25}=2.5\end{align*}

\begin{align*}\mbox{cos}\ \theta=\frac{\overrightarrow{MN} \times \overrightarrow{KL}} {|\overrightarrow{MN}| |\overrightarrow{KL}|}=\frac{3.375}{(2.25)(2.5)}=\frac{3.375}{5.625}=0.6\end{align*}

\begin{align*}\theta=\mbox{cos}^{-1}\ (-0.6)=53.1^\circ\end{align*}

By looking at the diagram, we can see that the angle between these two vectors is larger than 90o. Many calculators only give the smaller of the two angles between two lines. As you can see below, both θ and \begin{align*}\phi\end{align*} relate the blue line to the red line.

For our problem, the calculator returned a value of 53.1o. The actual angle between the two vectors is 180o - 53.1o = 126.9o when we take into account the directions of \begin{align*}\overrightarrow{MN}\end{align*} and \begin{align*}\overrightarrow{KL}\end{align*} .

6) The component form of the dot product is given by \begin{align*}\overrightarrow{w} \times \overrightarrow{h}=w_x h_x+w_yh_y+w_zh_z \end{align*}.

\begin{align*}\overrightarrow{w} \times \overrightarrow{h}=(85 \cdot 67)+(89 \cdot 70)+(91 \cdot 88)=5695+6320+8008=20023\end{align*}

Now we can find the angle between the two vectors using the other form of the dot-product equation: \begin{align*}\overrightarrow{A} \times \overrightarrow{B}= | A | | B |\ \mbox{cos}\ \theta\end{align*} , but first we need to determine the magnitudes of the two vectors using the Pythagorean Theorem.

\begin{align*}|\overrightarrow{w}|=\sqrt{w_x^2+w_y^2+w_z^2}=\sqrt{85^2+89^2+91^2}=\sqrt{7225+7921+8281}=\end{align*} \begin{align*}\sqrt{23427}=153.1\end{align*}

\begin{align*}|\overrightarrow{h}|=\sqrt{h_x^2+h_y^2+h_z^2}=\sqrt{67^2+70^2+88^2}=\sqrt{4489+4900+7744}=\end{align*} \begin{align*}\sqrt{17133}=130.9\end{align*}

\begin{align*}\mbox{cos}\ \theta=\frac{\overrightarrow{A} \times \overrightarrow{B}}{|A||B|}=\frac{20023}{(153.1)(130.9)} = 0.99911\end{align*}

\begin{align*}\theta=\mbox{cos}^{-1}\ 0.99911=2.42^\circ\end{align*}

### Practice

Find the Magnitude

1. What is the magnitude of \begin{align*}\left \langle -9, 3\right \rangle\end{align*}?
2. What is the magnitude of \begin{align*}\left \langle -3, 16 \right \rangle\end{align*}?
3. What is the magnitude of \begin{align*}\left \langle 6, 18\right \rangle\end{align*}?
4. What is the magnitude of \begin{align*}\left \langle 9, 10\right \rangle\end{align*}?

Find the Direction

1. What is the direction of \begin{align*}\left \langle -10, 10 \right \rangle\end{align*}?
2. What is the direction of \begin{align*}\left \langle 9, 17 \right \rangle\end{align*}?
3. What is the direction of \begin{align*}\left \langle 2, 17 \right \rangle\end{align*}?
4. What is the direction of \begin{align*}\left \langle -6, 12 \right \rangle\end{align*}?

Find: a) The dot product of Vectors A and B, and b) The measure of the angle found between the vectors, rounded to the nearest degree.

1. Vector A = \begin{align*} -2i + 5j \end{align*} and Vector B = \begin{align*} -3i + 5j \end{align*}
2. Vector A = \begin{align*}\left \langle 6, 13 \right \rangle\end{align*} and Vector B = \begin{align*}\left \langle 9, 10 \right \rangle\end{align*}
3. Vector A = \begin{align*}\left \langle 4, 5 \right \rangle\end{align*} and Vector B = \begin{align*}\left \langle -4, 8 \right \rangle\end{align*}
4. Vector A = \begin{align*} -7i + 4j \end{align*} and Vector B = \begin{align*}6i + 3j \end{align*}
5. Vector A = \begin{align*}\left \langle -9, 15 \right \rangle\end{align*} and Vector B = \begin{align*}\left \langle -7, 11 \right \rangle\end{align*}
6. Vector A = \begin{align*}\left \langle 1, 18 \right \rangle\end{align*} and Vector B = \begin{align*}\left \langle 1, 6 \right \rangle\end{align*}
7. Vector A = \begin{align*}4i + 11j \end{align*} and Vector B = \begin{align*}3i + 14j \end{align*}
8. Prove that \begin{align*}\overrightarrow{C} = 2\hat{i} + 5\hat{j} + 2\hat{k}\end{align*} and \begin{align*}\overrightarrow{D} = 3\hat{i} - 2\hat{j} + 2\hat{k}\end{align*}are perpendicular by showing their dot product is zero.
9. Explain why showing the dot product of two vectors is zero proves they are perpendicular.

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### Vocabulary Language: English

dot product

The dot product is also known as inner product or scalar product. The two forms of the dot product are $\vec{a} \cdot \vec{b} = \Big \| \vec{a}\Big \| \ \Big \| \vec{b}\Big \| \cos \theta$ and $\vec{a} \cdot \vec{b} = x_a x_b + y_a y_b$.

scalar product

A scalar product is the same as a dot product. The scalar product produces a number that can be interpreted to tell how much one vector goes in the direction of the other vector.

scalar projection

The scalar projection of a vector onto another vector is equal to the length of the projection of the first vector onto the second vector.

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