WEBVTT 00:00:19.863 --> 00:00:28.115 A unit vector is a vector with magnitude 1. We are particularly interested in two specific special unit 00:00:28.118 --> 00:00:38.635 vectors, i and j. Let's take a moment to talk about notation. A vector written by hand has an arrow over it 00:00:38.638 --> 00:00:44.643 to show that it is a vector. Note that when you see a vector typed, it's boldfaced. 00:00:44.646 --> 00:00:54.153 So the vector i is a unit vector travelling along the x-axis that is exactly 1 unit long. Component representation 00:00:54.156 --> 00:01:06.417 will be <1, 0>. The unit vector j is travelling along the y-axis, and its components are <0, 1>. 00:01:08.917 --> 00:01:16.426 Any vector v in standard position can be written in terms of the unit vectors i and j. 00:01:16.429 --> 00:01:24.424 Starting with a vector v in standard position with the terminal point (a, b), then this is the same as having 00:01:24.426 --> 00:01:45.422 a copies of i plus b copies of j. So we can write that the vector v with components a and b is the same 00:01:45.425 --> 00:02:01.472 as ai plus bj. Note that its magnitude will still be the square root of a squared plus b squared. 00:02:01.475 --> 00:02:09.220 This third representation of the vector v is particularly useful for algebraic manipulation. 00:02:11.468 --> 00:02:18.470 In this example, we're asked to calculate or find the unit vector u in the same direction as the vector v 00:02:18.473 --> 00:02:27.243 equals negative 3i plus 5j. We can find this unit vector by dividing our given vector by its magnitude, so 00:02:27.246 --> 00:02:50.270 let's calculate the magnitude of the vector first. The formula is the square root of a squared plus b squared. 00:02:50.273 --> 00:02:55.770 So the magnitude of our vector is the square root of 34. Of course you want to check to see if that can be 00:02:55.773 --> 00:03:05.019 simplified. Now we're ready to calculate the vector u. The unit vector in the same direction as v, having 00:03:05.022 --> 00:03:23.523 magnitude 1 to make it a unit vector, is going to be the vector v divided by its magnitude. We just calculated 00:03:23.526 --> 00:03:41.270 that the magnitude is the square root of 34, and we'll separate this so that we can see the components a and b. 00:03:43.772 --> 00:03:51.019 Let's look at this vector v drawn in standard position. Its terminal point is at (a,b), so I've written vector v 00:03:51.022 --> 00:03:59.520 in terms of its components and in ai plus bj form. Using this picture we can get several relationships 00:03:59.523 --> 00:04:12.770 between a, b, and the angle theta. First let's look at the cosine of theta. The cosine of theta is the 00:04:12.770 --> 00:04:20.359 adjacent side, which is a, over the hypotenuse of the triangle we see. How long is the hypotenuse? 00:04:20.362 --> 00:04:33.622 It's exactly the length of the vector, or its magnitude. Rearranging, we obtain this first relationship: 00:04:33.625 --> 00:04:43.634 a is the magnitude of the vector times the cosine of this angle theta. Now what about sine of theta? 00:04:43.637 --> 00:04:53.643 Sine theta is the opposite side, b, over the hypotenuse. The hypotenuse is the magnitude of our vector, and 00:04:53.646 --> 00:05:02.401 rearranging this one, we come up with the second relationship. Now let's talk a little bit more about theta. 00:05:02.404 --> 00:05:08.655 Theta is very special. It's called the direction angle of the vector. We're defining it to be a positive angle 00:05:08.658 --> 00:05:21.171 formed between the positive x-axis and the terminal side of our vector. Theta has to satisfy this equation. 00:05:21.174 --> 00:05:28.930 The tangent of theta, which should be the opposite over the adjacent, that would be b over a, where a is not 0. 00:05:28.933 --> 00:05:36.927 Notice that we've drawn our vector so that theta lies in quadrant one. We can easily extend this to any 00:05:36.930 --> 00:05:43.928 positive direction angle theta using our general angle definitions of the trig functions. 00:05:45.678 --> 00:05:51.937 In this example, we're asked to find the direction angle for this given vector. We'll start by drawing the vector 00:05:51.940 --> 00:06:08.179 in standard position. a is negative 6, so we go 6 units to the left; b is positive 6 square root of 3 so that is up 00:06:08.182 --> 00:06:18.728 6 square root of 3 units, and we draw our vector so that this is its terminal side. The direction angle theta 00:06:18.731 --> 00:06:26.477 will be this positive angle between the positive x-axis and our vector that was drawn in standard position. 00:06:26.480 --> 00:06:39.727 We know that this angle has to satisfy tangent of theta equals b over a, where a is not 0. Because the terminal 00:06:39.730 --> 00:06:56.244 side of our angle is not in quadrant one, we will use a reference angle to find theta. The reference angle is the 00:06:56.247 --> 00:07:03.727 inverse tangent of the absolute value of b over a. Remember that a reference angle needs to be positive and 00:07:03.730 --> 00:07:12.231 acute, so the absolute value is important here. 00:07:12.234 --> 00:07:25.226 b is 6 square root of 3; a is negative 6. Reducing this and taking the absolute value, we're looking for the 00:07:25.229 --> 00:07:33.728 inverse tangent of the square root of 3. This is a special angle. We should know the value of this. 00:07:33.731 --> 00:07:45.815 Inverse tangent of the square root of 3 is the angle pi over 3. Now use this to find theta. The reference angle is 00:07:45.818 --> 00:07:59.579 here, so theta is 180 degrees... so we should choose radians or degrees. 00:07:59.582 --> 00:08:12.079 Pi over 3 is 60 degrees, and then we'll write this one in degrees. 00:08:14.579 --> 00:08:22.077 In this example, we're given a direction angle of 330 degrees and a magnitude 10 of a vector v, and we're to 00:08:22.080 --> 00:08:31.579 write this in terms of i and j. That means we need to calculate a and b, the components of the vector. 00:08:31.582 --> 00:08:40.120 Recall that we know the relationships between a, b, theta, and the magnitude of the vector. 00:08:40.123 --> 00:08:52.378 There are two of those. The horizontal component a is the magnitude of the vector times the cosine of the 00:08:52.381 --> 00:09:03.880 direction angle, and then the vertical component of the vector is the magnitude of the vector times the sine of the 00:09:03.883 --> 00:09:19.377 direction angle. Substituting what we're given, magnitude 10, direction angle 330 degrees, 00:09:19.380 --> 00:09:27.916 we need to calculate these two values. Let's talk about 330 degrees for a moment. The terminal side of 00:09:27.919 --> 00:09:42.166 330 degrees is in quadrant four. The reference angle will be 30 degrees, and we know that in quadrant four 00:09:42.169 --> 00:09:55.166 the cosine of an angle is positive. So this will be 10 times the positive cosine of our reference angle. 00:09:55.169 --> 00:10:05.666 The cosine of 30 degrees is the square root of 3 over 2, and we're going to simplify that. 00:10:05.669 --> 00:10:15.215 Now on this one, 330 degrees, the terminal side is still in quadrant four. Reference angle is still 30 degrees, 00:10:15.218 --> 00:10:27.461 but in quadrant four the sine of an angle is negative, so this is negative sine of 30 degrees. The sine of 00:10:27.464 --> 00:10:49.745 30 degrees is 1/2. Simplifying that, we find that b is negative 5. We have to write this in ai plus bj form. 00:10:49.748 --> 00:10:55.748 Continue to practice these examples until you've learned to do them by yourself without help of any sort.