Math 1300 Distance: Tips for Assignment 2

Published: Wed, 01/30/13


 
Please note that my first review seminar for Math 1300 will be on Feb. 23.  Unfortunately, that means the seminar is the weekend at the end of the week-long Midterm Break, but it is out of my hands.  This seminar will cover lessons 1 to 8 of my book.  
 
For more info about the seminar, and to register if you have not already done so, please click this link:
Math 1300 Exam Prep Seminar 
 
I am also taking registrations for all my midterm exam prep seminars (Calculus, Linear Algebra, and Statistics).  Please click this link for more info and to register, if you are interested:
Grant's Exam Prep Seminars 
 
Did you read my tips on how to do well in Math 1300 Distance?  If not, here is a link to those important suggestions:
How to do Well in Math 1300 Distance 
 
Tips for Assignment 2
Even if you are not taking the distance course, I think it is very useful for all Math 1500 students to attempt these hand-in assignments.  In general, the assignments can be quite demanding and really force you to solidify your math skills.  However, the fact is that the distance course covers the topics in a different order from the classroom lecture sections, so I would not advise classroom students to look at this assignment until after the midterm exam.
 
Here is a link where you can download a copy of Assignment 2:
Math 1300 Distance Assignment 2 (Jan 2013) 
 
Don't have my book? You can download a sample containing two lessons (Lesson 1 and Lesson 2) here:
Grant's Tutoring Study Guides (Including Free Samples) 
 
You will need to study Lesson 10 (Equations of Lines and Planes) to prepare for assignment 2.
 
Question 1 is quite similar to the concepts I discuss in my question 4.
 
Question 2(b) 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.  The cross product of your two normal vectors will answer 2(c).
 
To get the distance they require in question 3(d), find the projection of the vector QP you found in part (c) onto the vector n that you found in part (a).  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 QP comes in.  If we have a point P on line 1 and a point Q on line 2, then we can visualize an arrow QP (or PQ) drawn from our line on the floor to the line on the ceiling.  But, don't visualize arrow QP 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 QP 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 QP 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.
 
Questions 4 to 6 are pretty much the same stuff again.