Planes and parametric equations

In summary, the conversation discusses a problem in linear algebra involving a plane and finding the z component of a specific point that lies on the plane. The problem is not in vector or parametric form, but it is easy to convert it to parametric form. The solution results in a line of infinite points with specific values for x, y, and z components. The problem also mentions finding a specific point that lies on the plane.
  • #1
grassstrip1
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OP warned about not using the homework template
Hi everyone! I'm having some issues with this problem for linear algebra. I understand parametric equations fairly but I'm confused about the unit vector notation

1) Consider the plane r(s,t)=2i + (t-s) j + (1+3s-5t) k find the z component of the point (2,-1, z0)

For what values of s and t is this the case?

I don't really know how to start the problem because it isn't in vector or parametric form like I'm used to.
 
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  • #2
grassstrip1 said:
Hi everyone! I'm having some issues with this problem for linear algebra. I understand parametric equations fairly but I'm confused about the unit vector notation

1) Consider the plane r(s,t)=2i + (t-s) j + (1+3s-5t) k find the z component of the point (2,-1, z0)

For what values of s and t is this the case?

I don't really know how to start the problem because it isn't in vector or parametric form like I'm used to.
It's easy enough to get from the vector form to the parametric form of this plane.
Here x = 2, y = t - s, and z = 1 + 3s - 5t, and you're given a point (2, -1, z0).
 
  • #3
I don't know what you are "used to" but it certainly is in "vector form" and, as Mark44 says, it is easy to convert to parametric form:
x= 2, y= t- s, z= 1+ 3s- 5t. In order to have [tex](x, y, z)= (2, -1, z_0)[/tex] you must have 2= 2, t- s= -1, and [tex]1+ 3s- 5t= z_0[/tex].

Perhaps it is the fact that there is not a single "unique" answer that is bothering you?

There are an infinite number of points, in fact an entire line, with x= 2, y= -1. From t- s= -1, we can get t= s+ 1 and so write [tex]z_0= 1+ 3s- 5(s+ 1)= 1+ 3s- 5s- 5= -4- 2s[/tex]. The set of such points consists of the line x= 2, y= -1, z= -4- 2s, for any s.
 
  • #4
Thank you for the replies! I left a little something out of the problem, it said find the z component so that it lies on the plane. Wouldn't that make it just one specific point?
 
  • #5
grassstrip1 said:
Thank you for the replies! I left a little something out of the problem, it said find the z component so that it lies on the plane. Wouldn't that make it just one specific point?
Work the problem through and see.
 

1. What are parametric equations and how are they used in planes?

Parametric equations are a set of equations that use parameters to define the coordinates of a point on a plane. In other words, they represent the x, y, and z coordinates of a point in terms of one or more independent variables. These equations are commonly used in planes to describe the motion of an object or to find the intersection points of two or more planes.

2. How do you convert a Cartesian equation to parametric form?

To convert a Cartesian equation to parametric form, you can set one of the variables (x, y, or z) equal to a parameter and then solve for the remaining variables in terms of that parameter. For example, if you have the equation x + 2y + 3z = 6, you can set x = t and solve for y and z in terms of t to get the parametric equations y = (6-t)/2 and z = (6-2t)/3.

3. What is the significance of the direction vector in parametric equations?

The direction vector in parametric equations represents the direction of the line or plane. It is a vector that is parallel to the line or plane and is used to determine the slope or direction of the line. In parametric equations, the direction vector is often represented as the coefficients of the independent variables (x, y, and z).

4. How do you find the distance between two planes using parametric equations?

To find the distance between two planes using parametric equations, you can first determine the normal vectors of each plane. Then, you can use the distance formula, d = |(a1x + b1y + c1z - d1) * n1| / |n1|, where a1, b1, c1, and d1 are the coefficients of the first plane, n1 is the normal vector of the first plane, and |n1| is the magnitude of the normal vector. This formula gives the shortest distance between the two planes.

5. What is the relationship between parametric equations and vectors?

Parametric equations are closely related to vectors. The direction vector in parametric equations is a vector that is parallel to the line or plane, and the position vector is a vector that represents the coordinates of a point on the line or plane. Additionally, the dot product of two direction vectors can be used to determine the angle between two lines or planes, and the cross product of two direction vectors can be used to find the normal vector of a plane.

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