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Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lessons 1 and 2: Did you read my tips on how to study and learn Math 1300? If not, here is a link to those important suggestions: Here is a link to the actual assignment, in case you don't have it handy: Study Lesson 10 (Lines and Planes) from my Linear Algebra & Vector Geometry book to prepare for this assignment.
Similar to my Lesson 10, question 4(b).
Part (a)
Similar to my Lesson 10, question 4(a).
Part (b)
Unfortunately, I do not discuss the symmetric equation of a line in my book (I
am surprised they mention it here). Like all equations for lines in 3-space, R^3, you need a point on the line p = (x0, y0, z0) and a vector parallel to the line, v = (a, b, c). Then the symmetric equations for the line are written:
Which is to say, subtract the coordinates of the point from (x,y,z) in the respective numerators, and divide by the parallel vector (a,b,c) in the respective denominators.
The symmetric equations are derived from the parametric equations for a line. Merely solve each parametric equation for t, and you end up with three different equations for t, which become the three symmetric
equations for the line. That is why the three parts are listed as equal to each other.
Note that the second line you are given in question 3 of the assignment is given in symmetric form.
Part (a)
You should be able to read the parallel vectors off for each line. If vectors are parallel (collinear), then their lines must be, too.
Part (b)
Your course
notes give you a formula for finding the distance between a point and a line that can be used here. Note that they give you four different formulas that could be used, two use the sine function and the other two use cross products. You can use whichever one you wish. The prudent student would just memorize one of the cross product formulas and use it whenever they need to get the distance between a point and a line. That is far simpler than using trig. You will note
the example they do uses one of the cross product formulas.
To find the distance between two lines, merely generate a point on one line (it should be easy for you to read
off one point on a line from the given equation for the line), and find the distance between that point and the other line.
Part (a)
Skew lines are lines that are not on the same plane (therefore they will never intersect), but are not parallel. For example, visualize a line drawn from east to west on the floor. Now visualize drawing a diagonal line on the
ceiling. Those lines are not parallel to each other, but they would never intersect either. Keep that visual image in your head as you do this problem.
To prove two lines are skew you have to do two things:
- Prove they are not parallel. (See question 3 above.)
- Prove the two lines do not intersect. (See Lesson 10, question 8. If two lines do not intersect there will be no solution to the
system. You will get a false result when trying to solve for s and t.)
Part (b)
Just use the formula for the cosine between two vectors from Lesson 9 using the two vectors read off the lines.
Part (c)
This is not an easy question! Here is a video I found on YouTube that is showing you what to do:
A Scotsman shows you how to solve a question like this. Click here.
Note that he writes his vectors in columns, rather than rows, but otherwise, you have to do the same thing with your two lines. First put the lines into parametric equations form.
Do they actually show you how
to do this in your course notes? I would appreciate it if someone could send me the course notes for this unit in the distance course. I would like to see if they delve into things this deeply, and, if so, direct you to whatever examples or notes they provide that may be relevant.
Part (a) If planes are parallel, they have the same normal vector. So that means they will have the same equation as the given plane, except the d value will be different. In other words, x+2y-3z=d. (Note that there is a
typo in their question. I am sure they meant that to be -3z, or else the two y terms would have been collected together.) Set up the equation to solve the distance between a point and a plane using this unknown d like I do in Lesson 10, question 10. You know the distance will be 2, so you can set up an equation to solve for d. As they suggest, you will get two solutions for
d.
Part (b) Just use the formula for the cosine between two vectors from Lesson 9 using the two vectors read off the planes.
Similar to my Lesson 10, question 9. Note that I use row-reduction to solve my problem (as taught in Lesson 2 of my book), but, technically, you aren't supposed to know that method yet. You can solve the problem using the methods I teach
in Lesson 1 of my book, too. See my Lesson 1, question 8 for a similar example. You will note that, by introducing a parameter t into the solution, you end up getting the parametric equations for a line.
Think about it. For the line to not intersect the plane, it must be parallel to the plane. How do you know if a line is parallel to a plane? I discuss this in Lesson 10. That will create an equation to solve for p. But!
Once you have solved p you should confirm that the line is not on the plane. Easiest way to do that is to show the given point on the line is not also on the plane.
Similar to my Lesson 10, question 8.
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