Question

Calculus III: Lines, planes, and surfaces in space.

Answer 1

@khoala4pham @B25 Find the parametric and symmetric equations of the line passing C(2, 3, 4) and perpendicular to 3x + 2y - z = 6.

Answer 2

Not sure how to use the information that the line is perpendicular to 3x + 2y - z = 6. I do not have a visualization of 3x + 2y - z = 6 either.

Answer 3

Have you learned the normal property of the equation of a plane? First off, I should start by saying, that 3x + 2y - z = 6 is a plane. Can you tell me something unique about this plane that comes immediately from looking at its equation perhaps something about the normal?

Answer 4

@khoala4pham Give me a moment to read through the section.

Answer 5

Sure thing.

Answer 6

Ok so the plane has a normal of\[n_p=<3,\ 2,\ -1>\]
What is the 6?

Answer 7

You are correct. The 6 doesn't matter though. In deriving the formula for the normal, the 6 "absorbs" all the other constants that you did not need. What is important is as you said, the normal is n = <3,2,-1>. Would you like me to clarify more on the 6 or is this enough? If not, then I'll continue with how to solve this problem.

Answer 8

@khoala4pham I plugged the planar equation into wolframalpha and changed from 6, to 0, to 10^(10000). Not sure how modifying it affects the graph, and I realize that it is not important to this problem, so we will move on and hopefully it will all come together.

Is this the proper parametrization?\[x=2+3t,\ y=3+2t,\ z=4-t\]

Answer 9

Yes. you are correct. like the last problem, you can find the symmetric equation easily. I will try to explain the 6, but it will take some time. Please have some patience. :)

Answer 10

@khoala4pham I am glad that is correct, it was a gander. Ok, so time to find the symmetric equations of the line.

Answer 11

Symmetric equations\[t=\frac{x-2}{3}=\frac{y-3}{2}=4-z\]

Answer 12

We assume that P(x,y,z) is an arbitrary point in the plane (of 3 space) and Q(a,b,c) be a fixed point on that same plane and that n is the normal to the plane given by n = <A,B,C>

It is obvious that we define the vector
r = <x,y,z> and the vector
b = <a,b,c>.

The line segment QP (watch the direction) is <x-a, y-b, z-c>.
By hypothesis, we know that n is tangent to the plane and using the property of the dot product, we know the dot product of N and QP is 0: two orthogonal vectors' dot product is 0.

You should also see that the vector QP is exactly equal to r-b. (For notation's sake, the dot product will be denoted as m*n, for assumed vectors m,n)

From the last two conclusions, we know:
n*(r-b) = 0. Expanding the dot product, we see
A(x-a) + B(y-b) + C(z-c) = 0. Voila. This form of a plane is known as the "point-normal form," the name is obvious.

Now where does that 6 come from? Well we can rewrite that point-normal form of a plane into Ax + By + Cz = D. where D = Aa + Bb + Cc. However, from our proof, D is the last thing we are concerned with and the most important conclusion is that D does NOT affect anything but the location of the plane.

Answer 13

@khoala4pham Thank you for your assistance. Last time I took Calculus III was a year ago, and your explanations are bring back clarity. If our parametrization is correct, we are going to move on to standard and general forms of planes. Again, I will tag you if you are able to join.

Answer 14

that's fine.