1. Planes and parametric equation

Hello everybody

We have four planes:

x + y + z = 0

2x - y + z= 1

x + y - 2z = 2

x - 2y + 2z = 1

And we know that they have a line L in common.

My question:

Can I find the parametric equation for L by solving this linear equation system ? Because the planes are just equations right ?

2. Re: Planes and parametric equation

Originally Posted by Tala
We have four planes:
x + y + z = 0
2x - y + z= 1
x + y - 2z = 2
x - 2y + 2z = 1

And we know that they have a line L in common.
My question:
Can I find the parametric equation for L by solving this linear equation system ? Because the planes are just equations right ?
You must find a point common to all four planes.
Then the cross product of two of the normals is the direction of the line.

3. Re: Planes and parametric equation

I'm supposed to use the method: Gauss Jordan elimination. So I write the A matrix (the coefficient matrix) and the b matrix and actually end up with a solution with one free parameter. Does this sound completely wrong ?

4. Re: Planes and parametric equation

What does "2x - y + z= 1 is the equation of a plane mean?" It means that if you have a point (a,b,c), then that point will be in that plane if and only if 2a - b + c = 1. (Ex: (1, 0, -1) and (2, 5, 2) are in the plane, but (3, 1, 1) is not.)

Aside: I used (a,b,c) there instead of (x, y, z) to make clear the distinction between the "equation of the plane" and "the (currently unknown) coordinates of a specific point on the plane". However, typically, we freely move between these two interpretations of the letters x, y, and z. This is exactly the same situation as the distinction between "the function" y = 2x -1, and the "a point on that line with coordinates (x,y), so that y = 2x - 1".

Now suppose a point (x, y, z) is on *all* of those four planes. (I could use (a, b, c), as above, but I'll do it the usual way.) Then that means that the numbers x, y, and z, satisfy the equation for each of those four planes. Thus x, y, and z satisfy all four of these conditions:
x + y + z = 0
2x - y + z = 1
x + y - 2z = 2
x - 2y + 2z = 1

Now, when we multiply both sides of any of those equations, we get a new equation that x, y, and z will satisfy.
If we add two of those equations - meaning add the two left hand sides and add the two right hand sides - then we'll get a new equation that x, y, and z still satisfy (sice we'll have added equal things to equal things, the results qill still be equal).
Thus you can proceed algebraically, or in terms of extended matricies and row reductions, but either way, the point is to solve that system of equations for x, y, and z as far as possible.

$\displaystyle \left(\begin{matrix}1 & 1 & 1 & | & 0 \\ 2 & -1 & 1 & | & 1 \\ 1 & 1 & -2 & | & 2 \\ 1 & -2 & 2 & | & 1 \end{matrix} \right)$

Clear 1st column:
$\displaystyle \left(\begin{matrix}1 & 1 & 1 & | & 0 \\ 0 & -3 & -1 & | & 1 \\ 0 & 0 & -3 & | & 2 \\ 0 & -3 & 1 & | & 1 \end{matrix} \right)$

Divide 2nd row by -3:
$\displaystyle \left(\begin{matrix}1 & 1 & 1 & | & 0 \\ 0 & 1 & 1/3 & | & -1/3 \\ 0 & 0 & -3 & | & 2 \\ 0 & -3 & 1 & | & 1 \end{matrix} \right)$

Clear 2nd column (including the 1st row):
$\displaystyle \left(\begin{matrix}1 & 0 & 2/3 & | & 1/3 \\ 0 & 1 & 1/3 & | & -1/3 \\ 0 & 0 & -3 & | & 2 \\ 0 & 0 & 2 & | & 0 \end{matrix} \right)$

Divide 3rd row by -3, 4th row by 2:
$\displaystyle \left(\begin{matrix}1 & 0 & 2/3 & | & -1/3 \\ 0 & 1 & 1/3 & | & -1/3 \\ 0 & 0 & 1 & | & -2/3 \\ 0 & 0 & 1 & | & 0 \end{matrix} \right)$

Now the last two equations are z = -2/3, and z = 0.

Thus this system has *no* solution. This is usually the case when you have more equations (here 4) than unknowns (here 3: x, y, z).

So either I made a mistake, or the problem you were given is incorrect - those four planes do *not* have a line L in common. Heck, they don't even have a single point in common.



Re: "Can I find the parametric equation for L by solving this linear equation system ? Because the planes are just equations right ?"

Unless I made a mistake, there is no such L. But it IS possible for 4 planes to intersect in a line - just not these 4 planes. If the numbers had been different, it might have worked out. And in that case, the answer to your question is YES.

SUPPOSE the result of the row reduction had been:

$\displaystyle \left(\begin{matrix}1 & 0 & -2 & | & -1 \\ 0 & 1 & 1 & | & 5 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{matrix} \right)$

Then that would say that: $\displaystyle x-2z = -1$ and $\displaystyle y+z = 5$.

That would mean that whenever (x, y, z) is on all four of those planes, that x, y, and z satisfy $\displaystyle x-2z = -1$ and $\displaystyle y+z = 5$.

And that's a parameteric representation of a line, with the z-coordinate as the free parameter:

$\displaystyle x = 2z-1$ and $\displaystyle y = -z+5$.

Maybe a better way to think about it is making t your free parameter, and then the parametric representation of the line is:

$\displaystyle x = 2t-1, y = -t+5, z = t$.

That's a line in 3-space = the set of points $\displaystyle \{ (2t-1, -t+5, t) \in \mathbb{R}^3 | t \in \mathbb{R} \}$.

To write that using vectors:

$\displaystyle (2t-1, -t+5, t) = (2t, -t, t) + (-1, 5, 0) = (2, -1, 1)t + (-1, 5, 0) = \vec{a}t + \vec{b}$,

where $\displaystyle \vec{a}= (2, -1, 1)$ and $\displaystyle \vec{b} = (-1, 5, 0)$.

So in terms of vectors, $\displaystyle L = \{\vec{a}t + \vec{b} \in \mathbb{R}^3 | t \in \mathbb{R} \}$.

FYI: **IF** this problem had row reduced as I described (it didn't), then some multiple of that vector $\displaystyle \vec{a}= (2, -1, 1)$ would'v been what you got when you did the cross product of the normals to the planes, as described in Plato's post.

How could Plato's approach have gotten a different vector (the same except for a multiple)? it's because when you're dealing with parametric or vector equations for a line (or plane), the representations ar not unique. That's because you can "rescale the free paramter". In this example, it might look like:

Have parametric for line L: $\displaystyle x = 2t-1, y = -t+5, z = t$

New representation, but still the same line L: $\displaystyle x = 20t-1, y = -10t+5, z = 10t$.

Yet another representation, but still the same line L (using new_t = -2(old_t)+8): $\displaystyle x = -4t+15, y = 2t-3, z = -2t+8$.

Have set description for line L: $\displaystyle \{ (2t-1, -t+5, t) \in \mathbb{R}^3 | t \in \mathbb{R} \}$.

New set description, but still the same line L: $\displaystyle \{ (-6t-1, 3t+5, -3t) \in \mathbb{R}^3 | t \in \mathbb{R} \}$.

Have vector description for line L: $\displaystyle \{\vec{a}t + \vec{b} \in \mathbb{R}^3 | t \in \mathbb{R} \}, \vec{a}= (2, -1, 1), \vec{b} = (-1, 5, 0)$.

New vector description, but still the same line L: $\displaystyle \{\vec{a_1}t + \vec{b_1} \in \mathbb{R}^3 | t \in \mathbb{R} \}, \vec{a_1}= (2\pi, -\pi, \pi), \vec{b_1} = (-1, 5, 0)$.

Notice for the vector description, it's the vector that changed and "drew in" the rescaling.

If we'd used new_t = -2(old_t)+8, then this would've happened:

$\displaystyle \vec{a}= (2, -1, 1), \vec{b} = (-1, 5, 0)$

$\displaystyle \vec{a}(-2t+8) + \vec{b} = -2\vec{a}t + 8\vec{a} + \vec{b} = (-2\vec{a})t + (8\vec{a} + \vec{b})$.

Since $\displaystyle \vec{a}= (2, -1, 1), \vec{b} = (-1, 5, 0)$, have $\displaystyle -2\vec{a} = (-4, 2, -2), (8\vec{a} + \vec{b}) = (15, 3, 8)$.

Thus $\displaystyle \vec{a}(-2t+8) + \vec{b} = \vec{a_2}t + \vec{b_2}$, where $\displaystyle \vec{a_2} = (-4, 2, -2), \vec{b_2} = (15, 3, 8)$.

So

New vector description, but still the same line L: $\displaystyle \{\vec{a_2}t + \vec{b_2} \in \mathbb{R}^3 | t \in \mathbb{R} \}, \vec{a_2}= (-4, 2, -2), \vec{b_2} = (15, 3, 8)$.

5. Re: Planes and parametric equation

This is exactly what I've done ! I'm so happy that it's correct.

6. Re: Planes and parametric equation

The simplest thing to do is to find a line common to two of the planes, then see if that line is in the other planes.
For example, adding the first two equations eliminates y so we have 3x+ 2z= 1 so that z= 1/2- (3/2)x. Putting that into the second equation, 2x- y+ 1/2- (3/2)x= 1 so that y= (1/2)x - 1/2. That gives the line x= t, y= (1/2)t- 1/2, z= 1/2- (3/2)t.
Now put that into the equations of the other two planes:
x + y - 2z = t+ (1/2)t- 1/2- 2(1/2- (3/2)t) = (9/2)t- 3/2, not 2 so there is NO line contained in these three planes (and so certainly not in all four).

x - 2y + 2z = 1