WEBVTT
00:00:01.020 --> 00:00:08.760
In this video, we will learn how to find direction angles and direction cosines for a given vector in space.
00:00:09.700 --> 00:00:13.670
We will begin by considering the three-dimensional coordinate grid.
00:00:14.960 --> 00:00:21.840
We know that in three-dimensional space, we have the π₯-, π¦-, and π§- or π§-axis.
00:00:22.600 --> 00:00:26.670
These are all perpendicular or at right angles to one another.
00:00:27.730 --> 00:00:38.110
Letβs assume we have the vector π― in the direction shown, where vector π― has components π― sub π₯, π― sub π¦, and π― sub π§.
00:00:38.110 --> 00:00:49.030
It is also important to recall that the unit vectors π’, π£, and π€ act in the π₯-, π¦-, and π§-direction, respectively.
00:00:49.840 --> 00:00:53.490
The first direction angle we need to be aware of is πΌ.
00:00:53.870 --> 00:00:57.480
This is the angle between the unit vector π’ and π―.
00:00:58.610 --> 00:01:04.920
Our second angle is the angle π½, which is the angle between the unit vector π£ and vector π―.
00:01:05.940 --> 00:01:12.130
Finally, we have the angle πΎ, which is the angle between the unit vector π€ and vector π―.
00:01:13.350 --> 00:01:17.770
These are the three direction angles πΌ, π½, and πΎ.
00:01:18.560 --> 00:01:21.580
Now letβs consider the direction cosines.
00:01:22.280 --> 00:01:29.630
We know that in right-angle trigonometry, the cos of any angle is equal to the adjacent over the hypotenuse.
00:01:30.430 --> 00:01:38.980
The direction cosines are simply the cos of our three angles, cos of πΌ, cos of π½, and cos of πΎ.
00:01:39.930 --> 00:01:48.880
cos of πΌ is equal to the π₯-component of the vector, in this case, π― sub π₯, divided by the magnitude of vector π―.
00:01:50.050 --> 00:01:57.030
In the same way, the cos of angle π½ is equal to π― sub π¦ divided by the magnitude of π―.
00:01:57.970 --> 00:02:05.430
And finally, the cos of angle πΎ is equal to π― sub π§ divided by the magnitude of π―.
00:02:06.330 --> 00:02:12.670
The cos of πΌ, cos of π½, and cos of πΎ are known as the direction cosines.
00:02:12.670 --> 00:02:22.360
By taking the inverse cos of both sides of these three equations, we get expressions for angle πΌ, π½, and πΎ.
00:02:23.310 --> 00:02:29.530
πΌ is equal to the inverse cos of π― sub π₯ over the magnitude of π―.
00:02:30.250 --> 00:02:37.050
Likewise, angle π½ is equal to the inverse cos of π― sub π¦ divided by the magnitude of π―.
00:02:37.920 --> 00:02:44.310
The angle πΎ is equal to the inverse cos of π― sub π§ over the magnitude of π―.
00:02:45.260 --> 00:03:01.130
We recall that if vector π― has components π― sub π₯, π― sub π¦, and π― sub π§, then the magnitude of vector π― is equal to the square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared.
00:03:02.490 --> 00:03:07.620
Note that this magnitude of a vector is also sometimes called the norm.
00:03:08.970 --> 00:03:14.710
In our first question, we will find a vector given its direction angles.
00:03:15.930 --> 00:03:27.430
Find vector π whose norm is 41 and whose direction angles are 135 degrees, 120 degrees, and 60 degrees.
00:03:28.300 --> 00:03:42.610
We know that the norm of a vector is its magnitude and that the direction vectors πΌ, π½, and πΎ are the angles between the unit vectors π’, π£, and π€ and the vector π―.
00:03:43.520 --> 00:04:08.900
We also know that the direction cosines are such that cos πΌ is equal to π― sub π₯ over the magnitude of π―, cos π½ is equal to π― sub π¦ over the magnitude of π―, and cos πΎ is equal to π― sub π§ over the magnitude of π―, where the vector π― has components π― sub π₯, π― sub π¦, and π― sub π§.
00:04:10.460 --> 00:04:20.940
Substituting in our values for πΌ and the magnitude, we see that the cos of 135 degrees is equal to π sub π₯ over 41.
00:04:21.790 --> 00:04:25.510
We can then multiply both sides of this equation by 41.
00:04:26.630 --> 00:04:31.530
π sub π₯ is equal to negative 41 root two over two.
00:04:32.420 --> 00:04:35.750
This is the π₯-component of our vector π.
00:04:36.650 --> 00:04:45.460
In the same way, considering angle π½, we have the cos of 120 degrees is equal to π sub π¦ over 41.
00:04:46.190 --> 00:04:55.720
Once again, we multiply both sides of the equation by 41 such that π sub π¦ is equal to negative 41 over two.
00:04:56.500 --> 00:05:04.720
Finally, we have the cos of 60 degrees, angle πΎ, is equal to π sub π§ divided by 41.
00:05:05.360 --> 00:05:16.180
π sub π§ is therefore equal to 41 multiplied by the cos of 60 degrees, which in turn is equal to 41 over two.
00:05:17.000 --> 00:05:27.580
This means that vector π has components negative 41 root two over two, negative 41 over two, and 41 over two.
00:05:28.450 --> 00:05:39.430
This is the vector whose magnitude or norm is 41 and direction angles are 135, 120, 60 degrees.
00:05:40.220 --> 00:05:46.790
Before looking at our next example, we will consider a formula that links the direction cosines.
00:05:48.480 --> 00:05:55.400
cos squared πΌ plus cos squared π½ plus cos squared πΎ is equal to one.
00:05:56.290 --> 00:06:01.180
From our knowledge of the direction cosines, letβs consider why this is true.
00:06:02.000 --> 00:06:06.990
We know that the cos of πΌ, π½, and πΎ are as shown.
00:06:07.800 --> 00:06:18.300
Squaring both sides of this equation, we see that cos squared πΌ is equal to π― sub π₯ squared divided by the magnitude of π― squared.
00:06:19.230 --> 00:06:35.280
Doing the same to our other two equations, we get cos squared π½ is equal to π― sub π¦ squared over the magnitude of π― squared and cos squared πΎ is equal to π― sub π§ squared over the magnitude of π― squared.
00:06:36.170 --> 00:06:46.080
We can now add the three expressions on the right-hand side as this would be equal to cos squared πΌ plus cos squared π½ plus cos squared πΎ.
00:06:46.910 --> 00:06:59.880
As all three fractions have the same denominator, this is equal to π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared all divided by the magnitude of π― squared.
00:07:00.880 --> 00:07:09.060
We know that the magnitude of π― is equal to the square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared.
00:07:09.750 --> 00:07:20.250
Squaring both sides of this equation gives us the magnitude of π― squared is equal to π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared.
00:07:21.040 --> 00:07:29.580
Substituting this into our expression gives us the magnitude of π― squared over the magnitude of π― squared.
00:07:30.340 --> 00:07:35.230
We know that dividing any scalar by itself gives us an answer of one.
00:07:35.930 --> 00:07:43.030
This proves that cos squared πΌ plus cos squared π½ plus cos squared πΎ is equal to one.
00:07:43.910 --> 00:07:47.740
We will now use this equation to help us solve a problem.
00:07:48.750 --> 00:07:56.560
Suppose that 31 degrees, 65 degrees, and π are the direction angles of a vector.
00:07:57.380 --> 00:08:01.170
Which of the following, to the nearest hundredth, is π?
00:08:02.130 --> 00:08:19.610
Is it (A) 72.88 degrees, (B) 84.00 degrees, (C) 85.03 degrees, or (D) 264.00 degrees?
00:08:20.320 --> 00:08:33.120
We recall that if the three direction angles of a vector are πΌ, π½, and πΎ, then cos squared πΌ plus cos squared π½ plus cos squared πΎ is equal to one.
00:08:33.120 --> 00:08:43.450
In this question, we will substitute πΌ equal to 31 degrees, π½ equal to 65 degrees, and πΎ equal to π.
00:08:43.550 --> 00:08:46.300
This gives us the equation shown.
00:08:46.430 --> 00:08:53.960
We recall that cos squared of 31 degrees is the same as cos of 31 degrees all squared.
00:08:53.960 --> 00:08:58.950
This is how we type this into a scientific calculator.
00:08:59.060 --> 00:09:10.070
cos squared 31 degrees plus cos squared 65 degrees is equal to 0.91334 and so on.
00:09:10.400 --> 00:09:23.620
We can then subtract this from both sides of our equation, giving us cos squared π is equal to 0.08665 and so on.
00:09:24.650 --> 00:09:27.640
We then square root both sides of this equation.
00:09:28.070 --> 00:09:33.860
cos π is equal to 0.29437 and so on.
00:09:34.580 --> 00:09:44.720
Finally, we take the inverse cos of both sides, such that π is equal to 72.8797 and so on.
00:09:45.600 --> 00:09:48.190
We want the answer to the nearest hundredth.
00:09:48.550 --> 00:09:52.580
This means that the nine in the thousandths column is the deciding number.
00:09:53.550 --> 00:09:57.980
Rounding up gives us 72.88 degrees.
00:09:58.140 --> 00:10:00.970
Therefore, the correct answer is option (A).
00:10:01.800 --> 00:10:11.400
The three direction angles of the vector are 61 degrees, 65 degrees, and 72.88 degrees.
00:10:12.640 --> 00:10:17.450
In our next question, we need to work out the direction cosines of a vector.
00:10:18.790 --> 00:10:25.670
Find the direction cosines of the vector π with components five, two, and eight.
00:10:26.700 --> 00:11:06.270
We recall that if vector π― has components π― sub π₯, π― sub π¦, and π― sub π§ and direction angles πΌ, π½, and πΎ, then the direction cosines cos of πΌ, cos of π½, and cos of πΎ are equal to π― sub π₯ over the magnitude of π―, π― sub π¦ over the magnitude of π―, and π― sub π§ over the magnitude of π―, respectively, where the magnitude of vector π― is equal to the square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared.
00:11:06.980 --> 00:11:12.650
In this question, we are told that vector π has components five, two, and eight.
00:11:13.340 --> 00:11:21.410
The magnitude of vector π is therefore equal to the square roots of five squared plus two squared plus eight squared.
00:11:21.980 --> 00:11:29.260
Five squared is equal to 25, two squared is equal to four, and eight squared is equal to 64.
00:11:30.110 --> 00:11:34.030
Summing these three values gives us an answer of 93.
00:11:34.410 --> 00:11:39.330
Therefore, the magnitude of vector π is the square root of 93.
00:11:40.140 --> 00:11:59.670
This means that the cos of angle πΌ is equal to five over the square root of 93, the cos of angle π½ is equal to two over the square root of 93, and, finally, the cos of angle πΎ is equal to eight over the square root of 93.
00:12:00.500 --> 00:12:12.090
The direction cosines of vector π are five over the square root of 93, two over the square root of 93, and eight over the square root of 93.
00:12:13.920 --> 00:12:20.690
In our final question, we will need to calculate the direction angles from a geometric problem.
00:12:22.010 --> 00:12:31.330
Find the measure of the direction angles of the vector π
represented by the given figure, correct to one decimal place.
00:12:32.310 --> 00:12:37.230
We will begin by writing the vector π
in terms of its three components.
00:12:38.240 --> 00:12:41.640
In the π₯-direction, we travel eight centimeters.
00:12:41.860 --> 00:12:45.090
Therefore, the π₯-component of the vector is eight.
00:12:46.160 --> 00:12:49.770
In the π¦-direction, we travel 19 centimeters.
00:12:49.940 --> 00:12:53.420
Therefore, the π¦-component of vector π
is 19.
00:12:54.420 --> 00:12:58.260
We travel nine centimeters in the π§- or π§-direction.
00:12:58.610 --> 00:13:01.240
Therefore, the π§-component is nine.
00:13:02.210 --> 00:13:06.210
Vector π
is equal to eight, 19, nine.
00:13:07.140 --> 00:13:23.780
The magnitude of vector π
is equal to the square root of eight squared plus 19 squared plus nine squared as we know that the magnitude of any vector is equal to the square root of the sum of the squares of the individual components.
00:13:24.500 --> 00:13:29.790
Eight squared plus 19 squared plus nine squared is equal to 506.
00:13:30.110 --> 00:13:35.380
Therefore, the magnitude of vector π
is the square root of 506.
00:13:36.300 --> 00:13:39.390
We are asked to calculate the direction angles.
00:13:39.900 --> 00:13:59.500
These are usually labeled πΌ, π½, and πΎ, where πΌ is the angle between the π₯-axis and the vector π
, π½ is the angle between the π¦-axis and the vector π
, and, finally, πΎ is the angle between the π§-axis and the vector π
.
00:14:00.200 --> 00:14:31.360
From our knowledge of the direction cosines, we know that πΌ is equal to the inverse cos of π
sub π₯ over the magnitude of π
, π½ is equal to the inverse cos of π
sub π¦ over the magnitude of π
, and πΎ is equal to the inverse cos of π
sub π§ over the magnitude of π
, where π
sub π₯, π
sub π¦, and π
sub π§ are the three components of vector π
.
00:14:32.420 --> 00:14:36.190
We will now clear some space to calculate these values.
00:14:37.280 --> 00:14:43.410
Angle πΌ is equal to the inverse cos of eight over the square root of 506.
00:14:43.410 --> 00:14:50.140
This is equal to 69.1671 and so on.
00:14:50.830 --> 00:14:57.000
Rounding to one decimal place, angle πΌ is equal to 69.2 degrees.
00:14:57.610 --> 00:15:03.920
π½ is equal to the inverse cos of 19 over the square root of 506.
00:15:04.520 --> 00:15:09.710
This is equal to 32.3652 and so on.
00:15:10.280 --> 00:15:15.420
Rounding this to one decimal place gives us 32.4 degrees.
00:15:16.410 --> 00:15:22.830
Angle πΎ is equal to the inverse cos of nine over the square root of 506.
00:15:23.320 --> 00:15:28.440
This is equal to 66.4156 and so on.
00:15:28.850 --> 00:15:34.340
To one decimal place, angle πΎ is 66.4 degrees.
00:15:35.010 --> 00:15:42.910
The angles πΌ, π½, and πΎ can also be written as π sub π₯, π sub π¦, and π sub π§.
00:15:42.910 --> 00:15:54.830
In this question, theyβre equal to 69.2 degrees, 32.4 degrees, and 66.4 degrees, respectively.
00:15:56.110 --> 00:15:59.740
We will know summarize the key points from this video.
00:16:01.030 --> 00:16:12.720
The direction angles β often denoted πΌ, π½, and πΎ β are the angles between a vector and the π₯-, π¦-, and π§- or π§-axes, respectively.
00:16:13.710 --> 00:16:47.320
The direction cosines of any vector are such that the cos of angle πΌ is equal to π― sub π₯ divided by the magnitude of vector π―, the cos of angle π½ is equal to π― sub π¦ over the magnitude of vector π―, and the cos of angle πΎ is equal to π― sub π§ over the magnitude of vector π―, where π― sub π₯, π― sub π¦, and π― sub π§ are the components of the vector in the π₯-, π¦-, and π§-directions.
00:16:48.030 --> 00:16:55.450
The magnitude of a vector is equal to the square root of the sum of the squares of the individual components.
00:16:56.240 --> 00:17:02.400
We also proved in this video that the sum of the squares of the direction cosines is equal to one.
00:17:03.490 --> 00:17:10.560
cos squared πΌ plus cos squared π½ plus cos squared πΎ is equal to one.
00:17:12.200 --> 00:17:20.430
We can use these equations to calculate the direction angles, direction cosines, or any other unknowns.