How Do You Determine the Vector Equation of a Plane Without a Z-Intercept?

[planauts]

Homework Statement

Find the vector equation of a plane that contains the following line L1: (x,y,z) = (1,2,3) + t(2,4,3) and does not have a z-intercept.

The Attempt at a Solution

This question was on the test and I think I might have gotten it wrong.
The correct answer is: (x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)

It makes sense to me now because (0,0,1) is parallel to the z-axis. Hence, it does not intercept the z-axis.

What I did was... think about the question in Cartesian Equation form.
Ax + By + Cz + D = 0 (where A,B,C are normal to the plane)

I realized that the co-efficient should be 0 since there is no z-intercept. So Ax + By + D = 0.
So the normal would be (A, B, 0).

Therefore,
(2,4,3) CROSS (x, y, z) = (A, B, 0) [where (x,y,z) is the second direction vector]

I did the cross product formula and ended up with:
(4z-3y, 3x-2z, 2y-4x) which should equal to (A, B, 0)
2y-4x = 0
2y=4x --> y = 2x and value of z doesn't matter

Therefore, (2,4,3) CROSS (x, 2x, 0) = (A, B, 0)
Let x = 1
So: (2,4,3) and (1,2,0) are the direction vectors.

So the equation of the plane is: (x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)

Could this be an acceptable answer?
Thanks

 

[CompuChip]

Unfortunately, if you set t = -1 and s = 1 (or t = 1, s = -3, and probably many other combinations) you will find a point on the z-axis, so this plane does intercept it.

 

[HallsofIvy]

Homework Statement

Find the vector equation of a plane that contains the following line L1: (x,y,z) = (1,2,3) + t(2,4,3) and does not have a z-intercept.

The Attempt at a Solution

This question was on the test and I think I might have gotten it wrong.
The correct answer is: (x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)

It makes sense to me now because (0,0,1) is parallel to the z-axis. Hence, it does not intercept the z-axis.

What I did was... think about the question in Cartesian Equation form.
Ax + By + Cz + D = 0 (where A,B,C are normal to the plane)

I realized that the co-efficient should be 0 since there is no z-intercept. So Ax + By + D = 0.
So the normal would be (A, B, 0).

At this point, a perfectly valid way to solve the probleme would be to note that when t=0, (x,y,z)= (1,2,3) and when t= 1, (x, y, z)= (1, 2, 3)+ (2, 4, 3)= (3, 6, 6) are on the given line and so on the plane. If the equation of the plane is Ax+ By+ D= 0, then A+ 2B+ D= 0 and 2A+ 4B+ D= 0 give two equations to solve (of course, any multiple of (A,B,D) is also a solution). Subtracting the two equations eliminates D and gives A+ 3B= 0 so that A= -3B. Taking B= 1 gives A= -3 and then -3+ 2+ D= 0 gives D= 1: -3x+ y= 1.

Therefore,
(2,4,3) CROSS (x, y, z) = (A, B, 0) [where (x,y,z) is the second direction vector]

I did the cross product formula and ended up with:
(4z-3y, 3x-2z, 2y-4x) which should equal to (A, B, 0)
2y-4x = 0
2y=4x --> y = 2x and value of z doesn't matter

Therefore, (2,4,3) CROSS (x, 2x, 0) = (A, B, 0)
Let x = 1
So: (2,4,3) and (1,2,0) are the direction vectors.

So the equation of the plane is: (x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)

Could this be an acceptable answer?
Thanks

 

[planauts]

At this point, a perfectly valid way to solve the probleme would be to note that when t=0, (x,y,z)= (1,2,3) and when t= 1, (x, y, z)= (1, 2, 3)+ (2, 4, 3)= (3, 6, 6) are on the given line and so on the plane. If the equation of the plane is Ax+ By+ D= 0, then A+ 2B+ D= 0 and 2A+ 4B+ D= 0 give two equations to solve (of course, any multiple of (A,B,D) is also a solution). Subtracting the two equations eliminates D and gives A+ 3B= 0 so that A= -3B. Taking B= 1 gives A= -3 and then -3+ 2+ D= 0 gives D= 1: -3x+ y= 1.

I need a vector equation.

So is my answer correct?
(x,y,z) = (1,2,3) + t(2,4,3) + s(1,2,0)

 

[HallsofIvy]

Same answer as you were given before: if you take t= -1/2, s= 0, you get (x,y,z)= (1- 1,2- 2, 3- 3/2)= (0, 0, 3/2) which is a "z-intercept". Your plane was to have NO z-intercept which means the z-axis must be parallel to it.

 

[planauts]

Oh, so this would be the correct answer, right?

(x,y,z) = (1,2,3) + t(2,4,3) + s(0,0,1)


Thanks for your help everyone,