WEBVTT
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A unit vector is a vector with magnitude 1. We are particularly interested in two specific special unit
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vectors, i and j. Let's take a moment to talk about notation. A vector written by hand has an arrow over it
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to show that it is a vector. Note that when you see a vector typed, it's boldfaced.
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So the vector i is a unit vector travelling along the x-axis that is exactly 1 unit long. Component representation
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will be <1, 0>. The unit vector j is travelling along the y-axis, and its components are <0, 1>.
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Any vector v in standard position can be written in terms of the unit vectors i and j.
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Starting with a vector v in standard position with the terminal point (a, b), then this is the same as having
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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
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as ai plus bj. Note that its magnitude will still be the square root of a squared plus b squared.
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This third representation of the vector v is particularly useful for algebraic manipulation.
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In this example, we're asked to calculate or find the unit vector u in the same direction as the vector v
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equals negative 3i plus 5j. We can find this unit vector by dividing our given vector by its magnitude, so
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let's calculate the magnitude of the vector first. The formula is the square root of a squared plus b squared.
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So the magnitude of our vector is the square root of 34. Of course you want to check to see if that can be
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simplified. Now we're ready to calculate the vector u. The unit vector in the same direction as v, having
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magnitude 1 to make it a unit vector, is going to be the vector v divided by its magnitude. We just calculated
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that the magnitude is the square root of 34, and we'll separate this so that we can see the components a and b.
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Let's look at this vector v drawn in standard position. Its terminal point is at (a,b), so I've written vector v
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in terms of its components and in ai plus bj form. Using this picture we can get several relationships
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between a, b, and the angle theta. First let's look at the cosine of theta. The cosine of theta is the
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adjacent side, which is a, over the hypotenuse of the triangle we see. How long is the hypotenuse?
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It's exactly the length of the vector, or its magnitude. Rearranging, we obtain this first relationship:
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a is the magnitude of the vector times the cosine of this angle theta. Now what about sine of theta?
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Sine theta is the opposite side, b, over the hypotenuse. The hypotenuse is the magnitude of our vector, and
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rearranging this one, we come up with the second relationship. Now let's talk a little bit more about theta.
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Theta is very special. It's called the direction angle of the vector. We're defining it to be a positive angle
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formed between the positive x-axis and the terminal side of our vector. Theta has to satisfy this equation.
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The tangent of theta, which should be the opposite over the adjacent, that would be b over a, where a is not 0.
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Notice that we've drawn our vector so that theta lies in quadrant one. We can easily extend this to any
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positive direction angle theta using our general angle definitions of the trig functions.
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In this example, we're asked to find the direction angle for this given vector. We'll start by drawing the vector
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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
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6 square root of 3 units, and we draw our vector so that this is its terminal side. The direction angle theta
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will be this positive angle between the positive x-axis and our vector that was drawn in standard position.
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We know that this angle has to satisfy tangent of theta equals b over a, where a is not 0. Because the terminal
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side of our angle is not in quadrant one, we will use a reference angle to find theta. The reference angle is the
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inverse tangent of the absolute value of b over a. Remember that a reference angle needs to be positive and
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acute, so the absolute value is important here.
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b is 6 square root of 3; a is negative 6. Reducing this and taking the absolute value, we're looking for the
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inverse tangent of the square root of 3. This is a special angle. We should know the value of this.
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Inverse tangent of the square root of 3 is the angle pi over 3. Now use this to find theta. The reference angle is
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here, so theta is 180 degrees... so we should choose radians or degrees.
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Pi over 3 is 60 degrees, and then we'll write this one in degrees.
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In this example, we're given a direction angle of 330 degrees and a magnitude 10 of a vector v, and we're to
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write this in terms of i and j. That means we need to calculate a and b, the components of the vector.
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Recall that we know the relationships between a, b, theta, and the magnitude of the vector.
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There are two of those. The horizontal component a is the magnitude of the vector times the cosine of the
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direction angle, and then the vertical component of the vector is the magnitude of the vector times the sine of the
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direction angle. Substituting what we're given, magnitude 10, direction angle 330 degrees,
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we need to calculate these two values. Let's talk about 330 degrees for a moment. The terminal side of
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330 degrees is in quadrant four. The reference angle will be 30 degrees, and we know that in quadrant four
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the cosine of an angle is positive. So this will be 10 times the positive cosine of our reference angle.
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The cosine of 30 degrees is the square root of 3 over 2, and we're going to simplify that.
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Now on this one, 330 degrees, the terminal side is still in quadrant four. Reference angle is still 30 degrees,
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but in quadrant four the sine of an angle is negative, so this is negative sine of 30 degrees. The sine of
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30 degrees is 1/2. Simplifying that, we find that b is negative 5. We have to write this in ai plus bj form.
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Continue to practice these examples until you've learned to do them by yourself without help of any sort.