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# 5.3: Dot Products

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Calculate the dot product of a pair of vectors.
• Determine the angle between a pair of vectors.
• Determine the vector and scalar projections of one vector onto another.

## Introduction

Previously, we discussed the multiplication of a vector by a scalar. There are actually two ways to multiply two vectors, both of which depend on the relative directions of the two vectors. One of the methods has its maximum when the two vectors are parallel; the other is maximized when the two vectors are perpendicular to one another. In this section we will look at the type of vector multiplication that gives a scalar value as the product. The Dot Product, also known as the scalar product, of two vectors will allow us to determine the angle between the two vectors and therefore will also help us to determine whether two vectors are parallel, perpendicular, or something in between.

## Dot Product

The dot product of two vector quantities is a scalar quantity which varies as the angle between the two vectors changes. The dot product is therefore sometimes referred to as the scalar product of the two vectors. The maximum value for the dot product occurs when the two vectors are parallel to one another, but when the two vectors are perpendicular to one another the value of the dot product is equal to 0. Furthermore, the dot product must satisfy several important properties of multiplication. The dot product must be commutative

\begin{align*}\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}\end{align*}

The dot product must also be distributive

\begin{align*}\overrightarrow{A} \times (\overrightarrow{B} + \overrightarrow{C}) = \overrightarrow{A} \times \overrightarrow{B} + \overrightarrow{A} \times \overrightarrow{C}\end{align*}

If we simply multiply the magnitudes of the two vectors together, we would fulfill the commutative property, but not the distributive property, since the magnitude

\begin{align*}| \overrightarrow{B} + \overrightarrow{C} |\ne|\overrightarrow{B}| + |\overrightarrow{C} |\end{align*}

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 vectors 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*}

Also, the dot product of any vector with the zero vector is equal to zero since the magnitude of the zero vector is itself equal to 0.

Now that we have two ways to compute the dot product, we can use these two methods to solve problems about vectors. The following examples illustrate one instance of this.

Example: (a) Calculate the dot product of the two vectors shown below. (b) Then 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: 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*}

Notice that these two vectors have these same magnitudes as the vectors in Example 1, but their dot product is not the same because they are placed at different angles relative to one another.

## Scalar Projections

A scalar projection of a vector onto a particular direction is given by the dot product of the vector with the unit vector for that direction. For example, the component notations for the vectors shown below are \begin{align*}AB =\left \langle {4,3} \right \rangle\end{align*} and \begin{align*}D =\left \langle {3,-1.25} \right \rangle\end{align*}.

The scalar projection of vector AB onto \begin{align*}\hat{x}\end{align*} is given by

\begin{align*}\overrightarrow{AB} \times \hat{x}=(4 \cdot 1)+(3 \cdot 0)+(0 \cdot 0)=4\end{align*}

The scalar projection of vector AB onto \begin{align*}\hat{y}\end{align*} is given by

\begin{align*}\overrightarrow{AB} \times \hat{y}=(4 \cdot 0)+(3 \cdot 1)+(0 \cdot 0)=3\end{align*}

And the scalar projection of vector AB onto \begin{align*}\hat{z}\end{align*} is given by

\begin{align*}\overrightarrow{AB} \times \hat{z}=(4 \cdot 0)+(3 \cdot 0)+(0 \cdot 1)=0\end{align*}

The scalar projections of AB onto the x and y directions are non-zero numbers because the vector is located in the x-y plane. The scalar projection of AB onto the z direction is equal to zero, because the z direction is perpendicular to AB.

Example: 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*}

## Vector Projections

The vector projection of a vector onto a given direction has a magnitude equal to the scalar projection. The direction of the vector projection is the same as the unit vector of that given direction. In a previous section, we saw that when a vector \begin{align*}\overrightarrow{v}\end{align*} is multiplied by a scalar s, its components are given by

\begin{align*}\overrightarrow{sv}=\left \langle sv_x,sv_y,sv_z \right \rangle\end{align*}

To calculate the vector projection of AB onto the direction of vector D, use the scalar projection calculated in the previous example and the unit vector \begin{align*}\overrightarrow{D}\end{align*}.

\begin{align*}(\overrightarrow{AB}\ \times \overrightarrow{D}) \overrightarrow{D}=\left \langle (1.23 \cdot 0.923),(1.23 \cdot -0.385)\right \rangle=\left \langle 1.135,-0.474 \right \rangle\end{align*}

## Lesson Summary

One of the two ways to multiply vector quantities is the Scalar Product. The scalar product, also known as the dot product, multiplies one vector by the component of the second vector which is parallel to the first. The result of a scalar product of two vectors is always a scalar number rather than a vector. There are two ways to calculate the dot product:\begin{align*}\overrightarrow{A}\ \times \overrightarrow{B}=|A||B|\ \mbox{cos}\ \theta\end{align*} and\begin{align*}\overrightarrow{A} \times \overrightarrow{B}= A_x B_x + A_y B_y + A_z B_z +...\end{align*}. These two versions of the dot product can be use to determine the angle between two vectors. The dot product can also be used to determine the scalar and vector projections of one vector in the direction of the other. The scalar projection of \begin{align*}\overrightarrow{A}\end{align*} onto the direction of \begin{align*} \overrightarrow{B}\end{align*} is given by \begin{align*}\overrightarrow{A} \times \overrightarrow{B}\end{align*} and the vector projection of \begin{align*}\overrightarrow{A}\end{align*} onto the direction of \begin{align*}\overrightarrow{B}\end{align*} is given by \begin{align*}(\overrightarrow{A} \times \overrightarrow{B}) \overrightarrow{B}\end{align*} , where \begin{align*}\overrightarrow{B}\end{align*} is the unit vector in the direction of \begin{align*}\overrightarrow{B}\end{align*}.

## Practice Problems

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 vector projection of \begin{align*}\overrightarrow{A}\end{align*} onto the direction of \begin{align*}\overrightarrow{B}\end{align*} and the vector projection of \begin{align*}\overrightarrow{B}\end{align*} onto the direction of \begin{align*}\overrightarrow{A}\end{align*}. \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*}.
4. 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 \langle 15,12.4,-3.7\right \rangle\end{align*} . Then determine the angle between the two vectors.
5. Determine the dot product of the following two vectors
6. Determine the scalar projection of the vector \begin{align*}\overrightarrow{R}=\left \langle 27,39,52 \right \rangle \end{align*} onto the direction of \begin{align*}\overrightarrow{T}=\left \langle 44,26,17 \right \rangle \end{align*} .
7. Determine the vector projection of vector \begin{align*}\overrightarrow{MN}\end{align*} onto the vector \begin{align*}\overrightarrow{KL}\end{align*} .
8. Determine the angle between the two vectors \begin{align*}\overrightarrow{MN}\end{align*} and \begin{align*}\overrightarrow{KL}\end{align*}.
9. 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. 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*} . 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. 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*}. 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. Determine the vector projection of \begin{align*}\overrightarrow{A}\end{align*} onto the direction of \begin{align*}\overrightarrow{B}\end{align*} and the vector projection of \begin{align*}\overrightarrow{B}\end{align*} onto the direction of \begin{align*}\overrightarrow{A}\end{align*} . \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*} . The vector projection of one vector onto the direction of another vector is given by \begin{align*}(\overrightarrow{A} \times \overrightarrow{B})\overrightarrow{B}\end{align*} , where\begin{align*}\overrightarrow{B}\end{align*} is the unit vector in the direction of \begin{align*}\overrightarrow{B}\end{align*} . Since it is a unit vector \begin{align*}\overrightarrow{B}\end{align*} has a magnitude of 1 and has the same direction as \begin{align*}\overrightarrow{B} ,\overrightarrow{B}=1\end{align*} @\begin{align*}-22^\circ\end{align*} . Therefore, \begin{align*}(\overrightarrow{A} \times \overrightarrow{B}) \overrightarrow{B}=(|\overrightarrow{A}||\overrightarrow{B}|\ \mbox{cos}\ \theta) \overrightarrow{B}=\left ((7)(1)\ \mbox{cos}\ 22\right )\end{align*} @ \begin{align*}-22^\circ = 6.49\end{align*} @ \begin{align*}-22^\circ\end{align*} The vector projection of one vector onto the direction of another vector is given by \begin{align*}(\overrightarrow{B} \times \overrightarrow{A})\overrightarrow{A}\end{align*} , where \begin{align*}\overrightarrow{A}\end{align*} is the unit vector in the direction of \begin{align*}\overrightarrow{A}\end{align*} . Since it is a unit vector \begin{align*}\overrightarrow{A}\end{align*} has a magnitude of 1 and has the same direction as \begin{align*}\overrightarrow{A}\end{align*} , \begin{align*}\overrightarrow{A}=1\end{align*} @ \begin{align*}0^\circ\end{align*} . Therefore, \begin{align*}(\overrightarrow{B} \times \overrightarrow{A})\overrightarrow{A}=(|\overrightarrow{B}||\overrightarrow{A}|\ \mbox{cos}\ \theta)\overrightarrow{A}=((4)(1)\mbox{cos}\ 22)\end{align*} @ \begin{align*}0^\circ = 3.71\end{align*} @ \begin{align*}0^\circ\end{align*}
4. 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. 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*}
5. Determine the dot product of the following two vectors. 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*}
6. Determine the scalar projection of the vector \begin{align*}\overrightarrow{R}=\left \langle 27,39,52 \right \rangle\end{align*} onto the direction of \begin{align*}\overrightarrow{T}=\left \langle 44,26,17\right \rangle\end{align*}. The scalar projection of one vector onto the direction of the other is the dot product of the first vector with the unit vector representing the direction of the second vector. To calculate the scalar projection, we need to determine the unit vector in the direction of vector \begin{align*}\overrightarrow{T}=\left \langle 44,26,17\right \rangle\end{align*} . Remember that a unit vector is equal to the ratio of the vector and its magnitude, therefore we first need to calculate the length of vector \begin{align*}\overrightarrow{T}\end{align*} . \begin{align*}|\overrightarrow{T}|=\sqrt{T_x^2+T_y^2+T_z^2}=\sqrt{(44)^2+(26)^2 +(17)^2}=\sqrt{1936+676+289}\end{align*} \begin{align*}=\sqrt{2901}=53.86\end{align*} \begin{align*}\overrightarrow{T}=\frac{\overrightarrow{T}}{|\overrightarrow{T}|}=\frac{\left \langle 44,26,17 \right \rangle}{53.86}=\left \langle \frac{44}{53.86},\frac{26}{53.86},\frac{17}{53.86} \right \rangle=\left \langle 0.8169,0.4827,0.3156 \right \rangle\end{align*} Now we can calculate the scalar projection of \begin{align*}\overrightarrow{R}\end{align*} onto \begin{align*}\overrightarrow{T}\end{align*} by calculating the dot product \begin{align*}\overrightarrow{R} \times \overrightarrow{T}=(27 \cdot 0.8169)+(39 \cdot 0.4827)+(52 \cdot 0.3156)=22.0563+18.6253\end{align*} \begin{align*}+ 16.4112=57.0928\end{align*}
7. Determine the vector projection of vector \begin{align*}\overrightarrow{MN}\end{align*} onto the vector \begin{align*}\overrightarrow{KL}\end{align*} . The vector progression of one vector onto a second vector is the multiplication of the dot products of the two vectors and the unit vector defining the direction of the second vector. In this case, \begin{align*}(\overrightarrow{MN} \times \overrightarrow{KL})\overrightarrow{KL}\end{align*} . First we need to identify the components of the two vectors by using the information given on the graph. In this case, \begin{align*}\overrightarrow{MN}=\left \langle -2.25,0.0\right \rangle\end{align*} and \begin{align*}\overrightarrow{KL}=\left \langle 1.5,2,0 \right \rangle\end{align*} . Then we need to determine the dot product of the two vectors. \begin{align*}\overrightarrow{MN} \times \overrightarrow{KL}=(MN)_x \ (MN)_x \ + \ (MN)_y \ (KL)_y \ + \ (MN)_z \ (KL)_z\end{align*} \begin{align*}= (2.25)(1.5) + (2)(0) + (0)(0) = 3.375\end{align*} We also need to determine the unit vector in the direction of \begin{align*}\overrightarrow{KL}\end{align*}.Remember that a unit vector is equal to the ratio of the vector and its magnitude, therefore we first need to calculate the length of vector \begin{align*}\overrightarrow{KL}\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*}\overrightarrow{KL}=\frac{\overrightarrow{KL}}{|\overrightarrow{KL}|}=\frac{\left \langle 1.5,2,0 \right \rangle}{2.5}=\left \langle \frac{1.5}{2.5},\frac{2}{2.5},\frac{0}{2.5} \right \rangle=\left \langle 0.6,0.8,0 \right \rangle\end{align*} Lastly, we multiply the dot product of the two vectors by this unit vector, \begin{align*}(\overrightarrow{MN} \times \overrightarrow{KL}) \overrightarrow{KL}=(3.375)\left \langle 0.6,0.8,0 \right \rangle=\left \langle 2.025,2.7,0 \right \rangle\end{align*}
8. 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\ (MN)_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*} .
9. 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. 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*}

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