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# 5.6: Vector Direction

Created by: CK-12

## Learning objectives

• Determine the direction angles which identify the direction of a particular vector.
• Determine the direction cosines which identify the direction of a particular vector.
• Determine the coefficients of a unit vector to represent a particular direction.

## Introduction

Throughout the previous sections we have used component notation to describe vectors. We have also used unit vectors to indicate a particular direction. In this section we will discuss the use of angles to specify a particular direction.

## Direction Angles

As we have seen before, the equation for a vector is given by

$\overrightarrow{p} = \left \langle P_x, P_y, P_z\right \rangle$

where Px, Py, and Pz are the x, y, and z coordinates of the vector obtained by projecting the vector onto the x, y, and z axes as shown below left.

The image above right, shows the angles between the position vector, $\overrightarrow{P}$, and the three axes: α is the angle between $\overrightarrow{P}$ and the x-axis, β is the angle between $\overrightarrow{P}$ and the y-axis, and γ is the angle between $\overrightarrow{P}$ and the z-axis.

The position vector, $\overrightarrow{P}$, and the unit vector, $\hat{x}$, define a plane shown in lavender below.

The direction angle, α, is the angle between $\overrightarrow{P}$ and $\hat{x}$ in the plane defined by the two vectors. The other plane shown in the diagram is the X-Y plane, which was included in the diagram to help you visualize orientation of the other plane.

In our discussion of the dot product, we saw that the dot product of two vectors can be given by $\overrightarrow{A} \times \overrightarrow{B} = | \overrightarrow{A} | | \overrightarrow{B} | \mbox{cos}\ \theta$

Therefore, we can calculate the angle between $\overrightarrow{P}$ and the unit vector $\hat{x}$.

$\alpha = \mbox{cos}^{-1} \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |}$

Similarly, the direction angles β and γ can be calculated using the equations

$\beta = \mbox{cos}^{-1} \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |}$ and $\gamma = \mbox{cos}^{-1} \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |}$

In some applications, such as astronomy and applied optics, the direction cosines are used at least as frequently as the directional angles themselves.

$\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |}, \mbox{cos}\ \beta \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |},$ and $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |}$

Example: Because airplanes move in three dimensions, air-craft ground crews can use direction cosines to identify their location at any moment. Rather than an arbitrary set of orthogonal x, y, and z axes, the position of an airplane is measured relative to the east, north, and zenith directions. (Zenith means top or upward.) At a particular moment, a small plane is 97 km east, 135 km north, and 7.5 km above its home airport. What are the directional cosines and directional angles for the position vector of the plane at that moment?

Solution: If we use the standard orientation of maps in the northern hemisphere, the x-direction corresponds to east, the y-direction to north, and the z-direction to the zenith. Therefore, the position vector of the plane can be written as $\overrightarrow{P} = \left \langle 297, 135, 7.5\right \rangle \mbox{km}$.

The direction cosines associated with this vector are given by

$\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}}$

$\mbox{cos}\ \alpha = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{297} {\sqrt{297^2 + 135^2 + 7.5^2}} = 0.582$

$\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}}$

$\mbox{cos}\ \beta = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{135} {\sqrt{297^2 + 135^2 + 7.5^2}} = 0.811$

$\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}}$

$\mbox{cos}\ \gamma = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{7500} {\sqrt{297^2 + 135^2 + 7.5^2}} = 0.045$

The associated direction angles are

α = cos-1 0.582 = 54.4o

β = cos-1 0.165 = 35.8o

γ = cos-1 0.045 = 87.4o

Example: In an upcoming episode of a crime drama, a swarm of special-effect insects bothers one of the investigators after the discovery of a murder victim in a drainage ditch. The animator uses position vectors to track the positions of the virtual pests with respect to an origin at the investigator’s head. One such insect is located at a point 33 cm in front, 52 cm to the left, and 18 cm below the tip of the investigator’s nose. What are the directional cosines for this insect?

Solution: As we look at our investigator, the +x direction is to her left, the +y direction is upward from her nose, and the +z direction is in front of her. The position vector of the midge can be written as $\overrightarrow{P} = \left \langle 33, 52, -18\right \rangle \mbox{cm}$.

The direction cosines associated with this vector are given by

$\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{P_x} {\sqrt{P^2_x + P_y^2 + P_z^2}}$

$\mbox{cos}\ \alpha = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{33} {\sqrt{33^2 + 52^2 + (-18)^2}} = 0.514$

$\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{P_y} {\sqrt{P^2_x + P_y^2 + P_z^2}}$

$\mbox{cos}\ \beta = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{52} {\sqrt{33^2 + 52^2 + (-18)^2}} = 0.810$

$\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{P_z} {\sqrt{P^2_x + P_y^2 + P_z^2}}$

$\mbox{cos}\ \gamma = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{-18} {\sqrt{33^2 + 52^2 + (-18)^2}} = 0.281$

## Pythagorean Property of Direction Cosines

An interesting property of direction cosines can be seen if we write the equations for the direction cosines in terms of the components of the position vector, Px, Py, and Pz and using the definition of the vector magnitude $| \overrightarrow{A} | = \sqrt{A^2_x + A^2_y + A^2_z}$

For example,

$\mbox{cos}\ \alpha = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}}$

The other two directional cosines can be rewritten similarly:

$\mbox{cos}\ \beta = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}}$ and $\mbox{cos}\ \gamma = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}}$

If we square both sides of all three equations and then sum them, we obtain

$\mbox{cos}^2 \alpha + \mbox{cos}^2 \beta + \mbox{cos}^2 \gamma = \frac{P^2_x} {P^2_x + P^2_y + P^2_z} + \frac{P^2_y} {P^2_x + P^2_y + P^2_z} + \frac{P^2_z} {P^2_x + P^2_y + P^2_z}$

which simplifies to

$\mbox{cos}^2 \alpha + \mbox{cos}^2 \beta + \mbox{cos}^2 \gamma = \frac{P^2_x + P^2_y + P^2_z} {P^2_x + P^2_y + P^2_z} = 1$

This is an important result because the definition of the unit vector stated that $| \hat{u} | = 1$

which also means that

ux2 + uy2 + uz2 = cos2α + cos2β + cos2γ

and that the components of the unit vectors correspond to the direction cosines.

Example: A local astronomer used direction cosines when programming the projector in a new planetarium dome. The projector itself sits at the center of the dome, 2.5 m above the floor. He wishes to project Mintaka, one of the stars in the belt of Orion, at a position 12 m south and 2.3 m east of the projector and 8.7 m above the floor. What is the equation of the directional unit vector that the astronomer must enter in to the projection computer?

Solution: We can use the same coordinate system that we used in Example 1 above: $\hat{x}$ = east, $\hat{y}$ = north, and $\hat{z}$ = upward. The projector itself is the origin. In such a coordinate system, the position vector for Mintaka becomes $\overrightarrow{P} = \left \langle 2.3, -12, (8.7 - 2.5)\right \rangle = \left \langle 2.3, -12, 6.2\right \rangle$. Notice that we did not use Mintaka’s position above the floor as the z-coordinate. Rather, since the projector is 2.5 m above the floor, we needed to use the difference between the ceiling height and the projector height. Use the component form of the directional cosine equation to calculate the three components of the unit vector.

$\mbox{cos}\ \alpha = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{2.3} {\sqrt{2.3^2 + (-12)^2 + 6.2^2}} = 0.168$

$\mbox{cos}\ \beta = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{-12} {\sqrt{2.3^2 + (-12)^2 + 6.2^2}} = -0.876$

$\mbox{cos}\ \gamma = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}} = \frac{6.2} {\sqrt{2.3^2 + (-12)^2 + 6.2^2}} = 0.453$

Therefore, the position unit vector is given by $\hat{u} = \left \langle 0.168, -0.876, 0.453\right \rangle$

## Lesson Summary

The direction of a particular vector can be identified using the components of the vector. The direction could also be defined using the directional angles between the vector and the coordinate axes. The dot product can be used to determine the cosine of the directional angles

$\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} {| \overrightarrow{P} |}, \mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} {| \overrightarrow{P} |},$ and $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} {| \overrightarrow{P} |}$

These angles are also related to the components of the unit vector corresponding to the direction of the vector $\overrightarrow{P}$

ux2 + uy2 + uz2 = cos2α + cos2β + cos2γ

## Practice Problems

1. Determine the components of the position vector $\overrightarrow{P} = \left \langle 2.4, 5.3, 1.8\right \rangle$ then determine the directional angles between this vector and the x-axis.
2. Determine the direction cosines for the vector $\overrightarrow{N} = \left \langle 8, 3, -5\right \rangle$
3. Determine the position vector for a small plane at the moment it is 2.5 km east, 8.8 km south, and 4.1 km above its home airport. Use a coordinate system where the x-direction corresponds to east, the y-direction to north, and the z-direction to the zenith. Then determine the directional cosines used by the air-traffic control personnel to identify the plane’s location.
4. Use the method of direction cosines to identify the unit vector having the same direction as the position vector $\overrightarrow{R} = \left \langle 781, 978, 1310\right \rangle$.
5. Determine the direction angles between the vector $\overrightarrow{p} = \left \langle 25, 8, 15\right \rangle$ and the coordinate axes.
6. In the crime drama scene described in the example, a swarm of special-effect midges bothers one of the investigators after the discovery of a murder victim in a drainage ditch. One midge is located at a point 7.2 cm in front, 22 cm to the right, and 14 cm above the tip of the investigator’s nose. What are the directional cosines used by the special-effects animator to identify this midge?

## Solutions

1. Determine the components of the position vector $\overrightarrow{P} = \left \langle 2.4, 5.3, 1.8\right \rangle$ then determine the directional angles between this vector and the x-axis. $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{2.4} {\sqrt{(2.4)^2 + (5.3)^2 + (1.8)^2}} = \frac{2.4} {\sqrt{5.76 + 28.09 + 3.24}} = \frac{2.4} {\sqrt{37.09}} = 0.394$ $\alpha = \mbox{cos}^{-1} 0.394 = 66.8$
2. Determine the direction cosines for the vector $\overrightarrow{N} = \left \langle 8, 3, -5\right \rangle$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{N} \times \hat{x}} { | \overrightarrow{N} |} = \frac{N_x} {\sqrt{N^2_x + N^2_y + N^2_z}}$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{N} \times \hat{x}} { | \overrightarrow{N} |} = \frac{N_x} {\sqrt{N^2_x + N^2_y + N^2_z}} = \frac{8} {\sqrt{(8)^2 + (3)^2 + (-5)^2}} = \frac{8} {\sqrt{64 + 9 + 25}} = 0.7213$ $\mbox{cos}\ \beta = \frac{\overrightarrow{N} \times \hat{y}} { | \overrightarrow{N} |} = \frac{N_y} {\sqrt{N^2_x + N^2_y + N^2_z}}$ $\mbox{cos}\ \beta = \frac{\overrightarrow{N} \times \hat{y}} { | \overrightarrow{N} |} = \frac{N_y} {\sqrt{N^2_x + N^2_y + N^2_z}} = \frac{3} {\sqrt{(8)^2 + (3)^2 + (-5)^2}} = \frac{3} {\sqrt{64 + 9 + 25}} = 0.3030$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{N} \times \hat{z}} { | \overrightarrow{N} |} = \frac{N_z} {\sqrt{N^2_x + N^2_y + N^2_z}}$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{N} \times \hat{z}} { | \overrightarrow{N} |} = \frac{N_z} {\sqrt{N^2_x + N^2_y + N^2_z}} = \frac{-5} {\sqrt{(8)^2 + (3)^2 + (-5)^2}} = \frac{-5} {\sqrt{64 + 9 + 25}} = -0.5051$
3. Determine the position vector for a small plane at the moment it is 2.5 km east, 8.8 km south, and 4.1 km above its home airport. Use a coordinate system where the x-direction corresponds to east, the y-direction to north, and the z-direction to the zenith. Then determine the directional cosines used by the air-traffic control personnel to identify the plane’s location. In this coordinate system, with an origin at the home airport, the position vector is given by $\overrightarrow{r} = \left \langle 2.5, -8.8, 4.1\right \rangle$ with units of kilometers. $\mbox{cos}\ \alpha = \frac{\overrightarrow{r} \times \hat{x}} { | \overrightarrow{r} |} = \frac{r_x} {\sqrt{r^2_x + r^2_y + r^2_z}}$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{r} \times \hat{x}} { | \overrightarrow{r} |} = \frac{r_x} {\sqrt{r^2_x + r^2_y + r^2_z}} = \frac{2.5} {\sqrt{(2.5)^2 + (-8.8)^2 + (4.1)^2}} = \frac{2.5} {\sqrt{100.5}} = 0.249$ $\mbox{cos}\ \beta = \frac{\overrightarrow{r} \times \hat{y}} { | \overrightarrow{r} |} = \frac{r_y} {\sqrt{r^2_x + r^2_y + r^2_z}}$ $\mbox{cos}\ \beta = \frac{\overrightarrow{r} \times \hat{y}} { | \overrightarrow{r} |} = \frac{r_y} {\sqrt{r^2_x + r^2_y + r^2_z}} = \frac{-8.8} {\sqrt{(2.5)^2 + (-8.8)^2 + (4.1)^2}} = \frac{-8.8} {\sqrt{100.5}} = -0.878$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{r} \times \hat{z}} { | \overrightarrow{r} |} = \frac{r_z} {\sqrt{r^2_x + r^2_y + r^2_z}}$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{r} \times \hat{z}} { | \overrightarrow{r} |} = \frac{r_z} {\sqrt{r^2_x + r^2_y + r^2_z}} = \frac{4.1} {\sqrt{(2.5)^2 + (-8.8)^2 + (4.1)^2}} = \frac{4.1} {\sqrt{100.5}} = 0.409$
4. Use the method of direction cosines to identify the unit vector having the same direction as the position vector $\overrightarrow{R} = \left \langle 781, 978, 1310\right \rangle$. The unit vector which has the same direction as this vector has the components $\hat{u} = \left \langle \mbox{cos}\ \alpha, \mbox{cos}\ \beta, \mbox{cos}\ \gamma\right \rangle$ where $\mbox{cos}\ \alpha = \frac{\overrightarrow{R} \times \hat{x}} { | \overrightarrow{R} |} = \frac{R_x} {\sqrt{R^2_x + R^2_y + R^2_z}}, \mbox{cos}\ \beta = \frac{\overrightarrow{R} \times \hat{y}} { | \overrightarrow{R} |} = \frac{R_y} {\sqrt{R^2_x + R^2_y + R^2_z}},$ and $\mbox{cos}\ \gamma = \frac{\overrightarrow{R} \times \hat{z}} { | \overrightarrow{R} |} = \frac{R_z} {\sqrt{R^2_x + R^2_y + R^2_z}}$. Once we find the three direction cosines, we have the components of the unit vector, $\mbox{cos}\ \alpha = \frac{\overrightarrow{R} \times \hat{x}} { | \overrightarrow{R} |} = \frac{791} {\sqrt{791^2 + 978^2 + 1310^2}} = \frac{791} {\sqrt{329826}} = \frac{791} {1816.11} = 0.436$ $\mbox{cos}\ \beta = \frac{\overrightarrow{R} \times \hat{y}} { | \overrightarrow{R} |} = \frac{978} {\sqrt{791^2 + 978^2 + 1310^2}} = \frac{978} {\sqrt{3298265}} = \frac{978} {1816.11} = 0.539$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{R} \times \hat{z}} { | \overrightarrow{R} |} = \frac{1310} {\sqrt{791^2 + 978^2 + 1310^2}} = \frac{1310} {\sqrt{3298265}} = \frac{1310} {1816.11} = 0.721$ $\hat{u} = \left \langle \mbox{cos}\ \alpha, \mbox{cos}\ \beta, \mbox{cos}\ \gamma\right \rangle = \left \langle 0.436, 0.539, 0.721\right \rangle$
5. Determine the direction angles between the vector $\overrightarrow{p} = \left \langle 25, 8, 15\right \rangle$ and the coordinate axes. $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{25} {\sqrt{(25)^2 + (8)^2 + (15)^2}} = \frac{25} {\sqrt{914}} = \frac{25} {30.23} = 0.827$ $\alpha = \mbox{cos}^{-1} 0.827 = 34.2^\circ$ $\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{8} {\sqrt{(25)^2 + (8)^2 + (15)^2}} = \frac{8} {\sqrt{914}} = \frac{8} {30.23} = 0.265$ $\beta = \mbox{cos}^{-1} 0.265 = 75.7^\circ$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{15} {\sqrt{(25)^2 + (8)^2 + (15)^2}} = \frac{15} {\sqrt{914}} = \frac{15} {30.23} = 0.496$ $\gamma = \mbox{cos}^{-1}\ 0.496 = 60.25^\circ$
6. In the crime drama scene described in the example, a swarm of special-effect midges bothers one of the investigators after the discovery of a murder victim in a drainage ditch. One midge is located at a point 7.2 cm in front, 22 cm to the right, and 14 cm above the tip of the investigator’s nose. What are the directional cosines used by the special-effects animator to identify this midge? As we look at our investigator, the +x direction is to her left, the +y direction is upward from her nose, and the +z direction is in front of her. The position vector of the midge can be written as $\overrightarrow{P} = \left \langle -22, 14, 7.2\right \rangle \mbox{cm}$. The direction cosines associated with this vector are given by $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{P_x} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \alpha = \frac{\overrightarrow{P} \times \hat{x}} { | \overrightarrow{P} |} = \frac{-22} {\sqrt{(-22)^2 + (14)^2 + (7.2)^2}} = \frac{-22} {\sqrt{731.84}} = \frac{-22} {27.05} = -0.813$ $\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{P_y} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \beta = \frac{\overrightarrow{P} \times \hat{y}} { | \overrightarrow{P} |} = \frac{14} {\sqrt{(-22)^2 + (14)^2 + (7.2)^2}} = \frac{14} {\sqrt{731.84}} = \frac{14} {27.05} = 0.518$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{P_z} {\sqrt{P^2_x + P^2_y + P^2_z}}$ $\mbox{cos}\ \gamma = \frac{\overrightarrow{P} \times \hat{z}} { | \overrightarrow{P} |} = \frac{7.2} {\sqrt{(-22)^2 + (14)^2 + (7.2)^2}} = \frac{7.2} {\sqrt{731.84}} = \frac{7.2} {27.05} = 0.266$

Feb 23, 2012

Dec 10, 2014