Equations of a Plane: Determining if a line lies within a plane

  • Thread starter Thread starter Buzzlastyear
  • Start date Start date
  • Tags Tags
    Line Plane
Click For Summary
SUMMARY

The discussion centers on determining whether the line defined by the equation (x, y, z) = (5, -4, 6) + u(1, 4, -1) lies within the plane represented by (x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1). The user initially finds that the point (5, -4, 6) lies within the plane but struggles to confirm if the entire line does. Through substitution and solving the parametric equations, it is concluded that the line does not lie in the plane, as the derived values do not satisfy the plane's equations consistently.

PREREQUISITES
  • Understanding of parametric equations in three-dimensional space
  • Knowledge of vector algebra and substitution methods
  • Familiarity with the concept of lines and planes in geometry
  • Ability to solve systems of equations
NEXT STEPS
  • Learn how to determine if a line intersects a plane using algebraic methods
  • Study the geometric interpretation of lines and planes in 3D space
  • Explore the use of vector equations in higher-dimensional geometry
  • Investigate the implications of parallel lines and planes in vector calculus
USEFUL FOR

Students studying geometry, particularly those focusing on vector equations and spatial relationships, as well as educators seeking to clarify concepts related to lines and planes in three-dimensional space.

Buzzlastyear
Messages
17
Reaction score
0

Homework Statement



Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Show algebraically.

Homework Equations


The Attempt at a Solution



I really have no idea where to start so I put the equations into parametric form to try to jog my brain but still got me nowhere!

(x,y,z)=(5,-4,6) + u(1,4,-1)
x=5+u
y=-4+4u
z=6-u

(x,y,z)=(3,0,2) + s(1,1,-1) + t(2,-1,1)
x=3+s+2t
y=s-t
z=2-s+t

EDIT: I was looking back in some of my notes and although there was nothing on how to determine if a line lies within a plane, I found how to determine if a point does. So I discovered that the point from the first vector equation (5,-4,6) does indeed lie within the plane (x,y,z)= (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). After discovering this however, i still don't see how i can use this information to see if the line (x,y,z)=(5,-4,6) + u(1,4,-1) lies within the plane

Not sure but maybe i have to sub in a value for s or t and solve the equations? i really have no leads so any help is very much appreciated thank you!

EDIT 2: I think i may have figured it out, here is what I've done:

substitute a value for u, ex. let u = 0

solve (x,y,z)=(5,-4,6)+0(1,4,-1)
therefore (5,-4,6) is a point when u=0

sub the values of x,y, and z into the parametric equations of the plane:

x=3+s+2t, y=s-t, z=2-s+t
6=3+s+2t, 0=s-t, 5=2-s+t
...s=t

sub s=t in 6=3+s+2t

6=3+t+2t
t=1

Sub t=1 in s=t

s=t
s=1

Now sub s=1 and t=1 into 5=2-s+t

LS=5 RS=2-s+t
RS=2-1+1
RS=2

Therefore the line (x,y,z)=(5,-4,6) + u(1,4,-1) does not lie in the plane??
 
Last edited:
Physics news on Phys.org
Find two points on your line and determine whether they satisfy the equation of your plane.
If both points on the line satisfy the plane equation, the line is in the plane.
If only one point satisfies the plane equation, the line intersects the plane but doesn't lie in it.
If neither point on the line satisifies the plane equation, the line is parallel to the plane and doesn't lie in it.
 
Thank you
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
2
Views
1K
Replies
5
Views
2K