Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
We’re talking about unit vectors. Let’s take a quick look at the unit circle. The unit circle is the circle x² plus y² equals 1. And remember the unit circle is centered at the origin and has a radius of 1. Suppose I draw a vector from the origin to any point on the unit circle. That would become a position vector for that point. And the position vectors for that point would be depending on this angle theta.
Observe that every one of these vectors is a unit vector. Why? Because the definition of a unit vector is a vector of length 1. And the radius of this circle is 1. So all of these vectors would be length 1. They are all unit vectors. Let’s use that to come up with some unit vectors in this problem. It says find the unit vector, u-hat in the direction, and first we’ll do theta equals pi over 3.
All you have to do to get a unit vector in the direction pi over 3, is write you have equals . The cosine of pi over 3 is 1/2 , and the sine of pi over 3 is root 3 over 2. There you are. There’s your unit vector in the direction pi over 3.
How about 3 pi over 4? Same thing; u equals cosine of 3 pi over 4, sine 3 pi over 4. You just have to remember what the cosine and sine of 3 pi over 4 are. Remember the reference angle would be pi over 4. The sine and cosine of pi over 4 are root 2 over 2.
In the second quadrant cosine will be negative, sine will be positive. So we get, negative root 2 over 2 for cosine, positive root 2 over 2 for sine. And that’s it, that’s your unit vector in the direction 3 pi over 4.
Let’s just do one more. Theta equals 3 pi over 2, that’s the straight downward direction. You could probably guess what the answer is but let’s use this formula, u-hat equals cosine 3 pi over 2, sine 3 pi over 2. The cosine of 3 pi over 2 is zero. And the sine of 3 pi over 2 is -1. There’s our unit vector, in the direction 3 pi over 2, <0, -1>.