Equation of plane given two parallel lines?

Click For Summary

Homework Help Overview

The discussion revolves around finding the equation of a plane that contains two given parallel lines represented by their equations. The problem involves understanding the relationships between the lines and the plane they define.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss converting the lines into parametric form and using the cross product to find a normal vector. There are attempts to identify points on the lines and calculate vectors between them. Some participants express confusion regarding specific calculations and the correct form of the equations.

Discussion Status

The discussion is ongoing, with participants actively exploring different approaches to the problem. There is a recognition of potential errors in calculations, and some participants are questioning the assumptions made in their attempts.

Contextual Notes

Participants note a correction regarding the original problem statement, which may affect the calculations and assumptions being discussed.

Isaac Wiebe
Messages
6
Reaction score
0

Homework Statement


Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3


Homework Equations



A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.


The Attempt at a Solution



Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

Since these lines are indeed parallel, we have to find another line or vector on the plane. Consider vector PQ, where point P (3, 0, 0) is on L1 and point Q (0.5, -2, 1) is on L2. Consequently, PQ is <-3.5, -2, 1>. The normal vector is found by taking the cross product of PQ and L1, and I am having a difficult time obtaining the correct answer, 24x - 13y + 34z = 72.
 
Physics news on Phys.org
Isaac Wiebe said:

Homework Statement


Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3


Homework Equations



A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.


The Attempt at a Solution



Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

I don't get that z.
 
Oops, the original problem read (1 - z) / 7, NOT (1 - z) / 3. My mistake!
 
Now check your calculations for PQ.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K