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Determining Cartesian Equation

  • Thread starter MPQC
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  • #1
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Homework Statement



"Determine the Cartesian equation of the plane that passes through the origin and contains the line [tex]\vec{}r[/tex] = (3,7,1) + t(2,2,3)

Homework Equations



Ax + By + Cz + D = 0

The Attempt at a Solution



Well. The way that I was taught to find the Cartesian Equation easily is to find the vector equation of the plane. Then using the two direction vectors, use the cross product of the two, and find the normal. Then the normal (x,y,z), are the same parts as the Cartesian equation, A, B, and C respectfully. Since I have the origin, I can plug in that point to find D, and I'll have the Cartesian equation.

But, I only have one direction vector in the equation I'm given. So I can't find the normal. So how can I solve it?
 

Answers and Replies

  • #2
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The point (3, 7, 1) is in your plane, and so is (0, 0, 0). These two points determine a vector. Use this vector and your direction vector to find a normal to the plane.
 
  • #3
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Wow, that was a lot easier then I thought it was going to be. Got the correct answer on my first shot at it. Thanks so much!

Sorry for another question

"Determine the vector, parametric, and Cartesian forms of the equation of the plane containing the lines L1: [tex]\vec{}r[/tex] = (3,-4,1) + s(1,-3,-5) and L2: [tex]\vec{}r2[/tex] = (7,-1,0) + t(2,-6,-10)"

Okay. So first thing I notice while looking at them is that they have the same direction vector. Simplifying the second equation brings it down to:

[tex]\vec{}r2[/tex] = (7,-1,0) + t(1,-3,-5)

So the only thing that is different between them is their origins. I looked at the answers, and they took both origins and found the vector that connected them, so in this case, it's (4,3,-1). But that's the part I don't understand - why did they do this?

The answer: [tex]\vec{}r[/tex] = (3,-4,1) + s(1,-3,-5) + t(4,3,-1)
 
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  • #4
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Since the direction vectors for L1 and L2 are parallel, you can't use them to get a normal to the plane, so you need to find another vector that's not parallel to the given direction vectors. An ideal choice is the vector from (3, -4, 1) to (7, -1, 0).
 
  • #5
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Determine the Cartesian equation of the plane that passes through the points (1,2,1) and (2,1,4) and is parallel to the x-axis. How the heck do you do this?
 
  • #6
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A(1,2,1)
B(2,1,4)

(Just as reference.)

Find the vector AB. Then, you know it's parallel to the x-axis, meaning your other vector is (1,0,0). You have your points, and two slopes. Just convert it normally. ;)
 
  • #7
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Great! Tks, now I get it.
BTW can you help on this one.

Find the vector equation of the plane that is parallel to the yz-plane and contains the point A(-1,2,1). Help please!
 
  • #8
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A normal to the plane is n = <1, 0, 0>. From the given point on the plane and a normal, it's easy to find the equation of the plane.
 
  • #9
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How exactly did you find that normal?
 
  • #10
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TKS Mark44.
 
  • #11
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BTW, I have yet more questions. First, I need to determine vector, parametric, and Cartesian equations for a plane that passes through the points P(1,-2,5) and Q(3,1,2) and is parallel to the line with the equation r = 2ti + (4t + 3)j + (t+1)k. I'd like to know what the line equation means, as I have never seen one like it and it is nowhere in my math book. Also, how can I find a plane parallel to that line equation? Any help would be highly appreciated.

Another question asks for an equation for the line that has the same x- and z-intercepts as the plane with the equation 2x + 5y - z + 7 = 0. I'd just like to know how you can determine the x and z-intercepts of the plane when I know that (2,5,-1) is the normal.

Help please!
 
  • #12
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How exactly did you find that normal?
It's given that the plane is parallel to the y-z plane, which makes it perpendicular to the x-axis.
 
  • #13
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BTW, I have yet more questions. First, I need to determine vector, parametric, and Cartesian equations for a plane that passes through the points P(1,-2,5) and Q(3,1,2) and is parallel to the line with the equation r = 2ti + (4t + 3)j + (t+1)k. I'd like to know what the line equation means, as I have never seen one like it and it is nowhere in my math book. Also, how can I find a plane parallel to that line equation? Any help would be highly appreciated.
The parametric equation of the line indicates that it goes the point (0, 3, 1) -- when t = 0 -- and the line is in the direction of the vector <2, 4, 1>. To get to any point on that line, start with the vector from the origin to the point P(0, 3, 1), and add the appropriate scalar multiple of the vector <2, 4, 1>.

There are an infinite number of planes that are parallel to this line.
Another question asks for an equation for the line that has the same x- and z-intercepts as the plane with the equation 2x + 5y - z + 7 = 0. I'd just like to know how you can determine the x and z-intercepts of the plane when I know that (2,5,-1) is the normal.

Help please!
To find the x-intercept, set y and z = 0 in your plane equation. To find the z-intercept, set x and y = 0 in the plane equation. After you find the two intercepts, you can find the equation of the line. Keep in mind what I said about the other problem in this post.

I don't believe that the normal to the plane enters into things.
 
  • #14
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The parametric equation of the line indicates that it goes the point (0, 3, 1) -- when t = 0 -- and the line is in the direction of the vector <2, 4, 1>. To get to any point on that line, start with the vector from the origin to the point P(0, 3, 1), and add the appropriate scalar multiple of the vector <2, 4, 1>.
How did you get that? The line equation uses addition, which makes it look like a vector equation. How do you make that equation look like a normal vector equation? What do the unit vectors i, j, and k play into all of this or do they really not matter or do they simply signify x, y, and z? I sort of get it and sort of don't at the same time.

Please help!
 
  • #15
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Also, how do you then find a plane parallel to it?
 
  • #16
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A normal to the plane is n = <1, 0, 0>. From the given point on the plane and a normal, it's easy to find the equation of the plane.
Ok, so I got the Cartesian equation of x + 1 = 0. But I need the vector and parametric equations for it. My teacher says that you can get a vector equation from a Cartesian equation of a plane by finding three points on the plane and finding two direction vectors with them. Fine. But there are no y and z components on the plane equation because it's really 1x + 0y + 0z + 1 = 0, so does that mean that pretty much any point is on the plane? If so, do you just take random points like (-1,2,3), (-1,1,1), and (-1,1,0) and use those? This is my point of confusion here.

There is another question that has me confused a bit. It says: The two lines L1: r = (-1,1,0)+ s(2,1,-1) and L2: r = (2,1,2) + t(2,1,-1) are parallel but do not coincide. The point A(5,4,-3) is on L1. Determine the coordinates of a point B on L2 such that vector AB is perpendicular to L2.

I'm not sure where to start on this one. I know that eventually AB dot (2,1,-1) will equal 0, but I'm not really sure what to do with this question. You've been really helpful so far. Thanks so far!
 
  • #17
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The parametric equation of the line indicates that it goes the point (0, 3, 1) -- when t = 0 -- and the line is in the direction of the vector <2, 4, 1>. To get to any point on that line, start with the vector from the origin to the point P(0, 3, 1), and add the appropriate scalar multiple of the vector <2, 4, 1>.
How did you get that? The line equation uses addition, which makes it look like a vector equation. How do you make that equation look like a normal vector equation? What do the unit vectors i, j, and k play into all of this or do they really not matter or do they simply signify x, y, and z? I sort of get it and sort of don't at the same time.
I usually discard the i, j, and k unit vectors, and write a vector as a list using < > notation, which is shorter. And, yes, these unit vectors are vectors in the direction of the x-, y-, and z-axis, respectively. If you keep track of things, you don't need them.

Your line equation was r = 2ti + (4t + 3)j + (t+1)k, and that really should be
r(t) = 2ti + (4t + 3)j + (t+1)k

Getting rid of the i, j, and k, we have r(t) = <0, 3, 1> + t<2, 4, 1>
r(0) = <0, 3, 1> + 0<2, 4, 1> = <0, 3, 1>

r(t) is what is called a vector-valued function of one parameter. Each value of t gives you a new vector, which is the vector sum of <0, 3, 1> and some multiple of <2, 4, 1>.
 
  • #18
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Ok, so I got the Cartesian equation of x + 1 = 0. But I need the vector and parametric equations for it. My teacher says that you can get a vector equation from a Cartesian equation of a plane by finding three points on the plane and finding two direction vectors with them. Fine. But there are no y and z components on the plane equation because it's really 1x + 0y + 0z + 1 = 0, so does that mean that pretty much any point is on the plane? If so, do you just take random points like (-1,2,3), (-1,1,1), and (-1,1,0) and use those? This is my point of confusion here.
The equation x + 1 = 0 is correct, but some of your thinking is not. y and z are arbitrary in this equation, but that doesn't mean that just any old point is on the plane. Only those points for which x = -1 are on the plane, so (-1, 2, 3), (-1, 1, 1), and (-1, 1, 0) are on the plane, while, for example (2, 0, 0) is not on the plane.

Your parametric equation will necessarily involve two parameters, say s and t. r = r(s, t) is a vector-valued function of two parameters.

If you know that the plane goes through P(a, b, c) and the plane contains two nonparallel vectors u = <u1, u2, u3> and v = <v1, v2, v3>, the parameteric equation will be
r(s, t) = <a, b, c> + su + tv.

Geometrically, what this means is that any point in the plane is the vector sum of a vector from the origin to point P, plus some linear combination (the sum of multiples) of the two direction vectors.
There is another question that has me confused a bit. It says: The two lines L1: r = (-1,1,0)+ s(2,1,-1) and L2: r = (2,1,2) + t(2,1,-1) are parallel but do not coincide. The point A(5,4,-3) is on L1. Determine the coordinates of a point B on L2 such that vector AB is perpendicular to L2.

I'm not sure where to start on this one. I know that eventually AB dot (2,1,-1) will equal 0, but I'm not really sure what to do with this question. You've been really helpful so far. Thanks so far!
How about starting a new thread for this problem? You really should limit a thread to one problem.
 
  • #19
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How about starting a new thread for this problem? You really should limit a thread to one problem.
How do I do that? And I have more questions. So I should make more than one thread?
 
  • #20
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When you open the Calculus & Beyond section, there's a button at the top left that says NEW TOPIC. Click that button to start a new topic (thread). And yes, you should start a new topic for each question. Jumping into the middle of an existing thread with a new question is considered "highjacking" the thread.
 
  • #21
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TKs. Soory for the Highjacking. New at this. You are a great help.
I will start new threads.
 

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