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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.
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.
Be careful! That second line appears to be in symmetric form,
but it is not! It is in no form known to man for a line, and tells you nothing as is. Set each part of that equation equal to t and proceed to solve for x, y and z in each case, to convert the line to parametric equations. Now you will be ready to proceed. Actually, since t is already being used as a parameter in the first line, you should really use s for a parameter in this second line.
Clarifying, take the first part of the second
equation and set it equal to s. Now isolate x in that equation. Take the second part and set it equal to s. Now isolate y in that equation. Repeat for the z part, and you will now have parametric equations for x, y and z.
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.
Easier approach to do part (c) - Find a vector
normal to both lines (cross product will help here). Let's call that vector n.
- Choose a point P on the first line and a point Q on the second line (you should be able to read a point off each line easily), then compute arrow PQ (or arrow QP is just as good). Let's call that vector v.
- Now, find the projection of v onto n. (You are finding the "shadow" of the arrow connecting P and Q cast onto a
vector that is at right angles to both lines.)
- Find the norm or length of that projection vector, and you have found the shortest distance between the two lines.
Even easier way to do part (c)
This is a simplification of the way I just described, but I don't know if you want to do this since you wouldn't be giving much explanation as to why it works. Compute vectors n and v, as I explain
above. Then, the distance between the two lines is simply v dot n divided by the norm of n.
Try that out on the Scotsman's example above. You should get the same answer. Note 3 divide by square root 2 is equivalent to 3/2 times square root 2. Just multiply both top and bottom by square root 2 to prove that.
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, 2x+y-4z=d. 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.
Like my Lesson 10, question 5. Just do the question with the p. However, for there to be no point of intersection, there must be no solution to the equation.
When finding the point of intersection between a line
and a plane, three things can happen:
- You get a solution for t. That value of t gives you a point of intersection.
- All the t terms cancel out, and you end up getting a true equation with no t at all. Like 6=6, or 10=10, or 0=0. That proves there are infinite points of intersection. The line is actually on the plane.
- All the t terms cancel out, and you end up getting a false
equation with no t at all. Like 6=3, or 10=19, or 0=8. That proves there are no points of intersection. The line is actually parallel to the plane.
So you must establish a value of p to cause case 3. Set the t terms in your intersection equation equal to 0 and solve for p then confirm that value of p causes case 3.
Alternative Method:
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. Or, now that you have solved
p, determine the actual equation of the line and now confirm that the line and plane have no point of intersection as I discussed above.
Similar to my Lesson 10, question 8.
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