Math 1300: Tips for Distance Assignment 2
Published: Thu, 10/11/12
Here are some tips to help you with Assignment 2 if you are taking Math 1300 via distance/online.
You will need to study Lesson 10 (Equations of Lines and Planes) to prepare for assignment 2.
Questions 1 and 2 are quite similar to the concepts I discuss in my question 4.
Question 3 is quite straightforward if you have studied my lesson. Note that you will need the distance between a point and a plane formula that I discuss in my question 10.
To get the distance they require in question 4(e), find the projection of the vector PQ you found in part (d) onto the vector n that you found in part (b). The length of that projection vector is the distance you require.
Why? Visualize two skew lines. For example, imagine a line running east-west on your floor. Imagine the second line to be running northeast-southwest along your ceiling (assuming the ceiling is parallel to the floor). There is an example of two skew lines. They are not parallel to each other, but they also never intersect.
Now visualize a vector normal to those two lines, n. That would be like a vertical pole rising from the line on the floor through the line on the ceiling. The problem is that n may actually be too short to touch both lines, or too long (going way beyond the two lines). That is why n will not tell you the distance between the two lines. However, if we could find a vector collinear to n that is exactly the same length as the distance between the two lines, then that vector will give us the distance we desire.
That is where arrow PQ comes in. If we have a point P on line 1 and a point Q on line 2, then we can visualize an arrow PQ drawn from our line on the floor to the line on the ceiling. But, don't visualize arrow PQ as a vertical arrow going straight up in the air. We don't know if that is true. Deliberately visualize P and Q in such a way that you are drawing a diagonal line from P on the floor to Q on the ceiling.
If we now compute the projection of PQ onto n that will give us a vector that is collinear with n but exactly the length we need (because that vector is the vertical leg of a right triangle with PQ as its hypotenuse). Find the length of that projection vector and we have found this distance.
By the way, this is the origin of the distance between a point and a plane formula.
Question 5 is pretty standard stuff again.
Question 6(d) is just a matter of using your cosine of the angle between two vectors formula from Lesson 9. Merely find the cosine of the angle between the two normal vectors.