Proving the line lies on the plane

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In summary, the question is whether the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lies in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). To answer this, we can simplify the equations and show that the point (5, -4, 6) lies on the plane by proving it can be written as a linear combination of the vectors (1,1,-1) and (2, -1, 1). Similarly, we can show that the vector (1,4
  • #1
soulja101
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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)? Justify your answer algebraically.


Homework Equations




The Attempt at a Solution


I started by getting the parametric equation of (x, y, z) = (5, -4, 6) + u(1,4,-1)
x=5+u
y=-4+4u
z=6-u
I then subbed in u=0 to get a set of points
x=5
y=-4
z=6

I then got the parametric equation for (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)
x=3+s+2t
y=0+s-t
z=2-s+t

I decided to use the points (5,-4,6)
 
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  • #2
Hi soulja101! :smile:
soulja101 said:
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)? Justify your answer algebraically.

I started by getting the parametric equation of (x, y, z) = (5, -4, 6) + u(1,4,-1)
x=5+u
y=-4+4u
z=6-u
I then subbed in u=0 to get a set of points
x=5
y=-4
z=6

This is very long-winded :rolleyes:

you could just put u = 0 in (x, y, z) = (5, -4, 6) + u(1,4,-1), giving (5, -4, 6) immediately :wink:
I then got the parametric equation for (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)
x=3+s+2t
y=0+s-t
z=2-s+t

why make it so complicated?

all you have to prove is that (5, -4, 6) minus (3, 0, 2) is a linear combination of (1,1,-1) and (2, -1, 1), and then the same for (1,4,-1) :smile:
 
  • #3
to see if they fit the equation of the plane. So, I plugged in x=5, y=-4, and z=6 into the parametric equations for the plane (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1).

After simplifying, I got:
5=3+s+2t
-4=0+s-t
6=2-s+t

Solving this system of equations, I got s=2 and t=0. Substituting these values back into the parametric equation for the plane, I got:
(x, y, z) = (3, 0, 2) + 2(1,1,-1) + 0(2, -1, 1)
(x, y, z) = (5, -4, 6)

Since the point (5,-4,6) satisfies the equation of the plane, we can conclude that the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) does lie on the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). This is because the parametric equation of the line is a subset of the parametric equation of the plane, meaning that all points on the line will also satisfy the equation of the plane.
 

1. How can you prove that a line lies on a plane?

To prove that a line lies on a plane, you can use the slope-intercept form of the line equation and the point-slope form of the plane equation. If the coordinates of the points on the line satisfy the plane equation, then the line lies on the plane.

2. What is the difference between a line and a plane?

A line is a one-dimensional geometric figure that extends infinitely in both directions, while a plane is a two-dimensional surface that extends infinitely in all directions. In other words, a line has only length, while a plane has both length and width.

3. What are some real-life applications of proving a line lies on a plane?

Proving a line lies on a plane is essential in many fields, including geometry, engineering, and physics. It is used to determine the intersection of two or more planes, to find the shortest path between two points on a plane, and to solve problems involving angles and distances on a plane.

4. Can a line lie on more than one plane?

Yes, a line can lie on more than one plane. In fact, if a line is parallel to a plane, it lies on an infinite number of planes that are parallel to the original plane.

5. What happens if a line does not lie on a plane?

If a line does not lie on a plane, it means that the coordinates of the points on the line do not satisfy the plane equation. In this case, the line and the plane do not intersect, and the line is said to be skew to the plane.

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