Linear algebra - linear equation for a plane

[Rectifier]

The problem
I am trying to write the equation for the plane on the following form $ax + by + cz + d = 0$

$$
\begin{cases}
x = 1 + s - t \\
y = 2 - s \\
z = -1 + 2s
\end{cases}
$$

The attempt
$s, t$ are the parameters for the two directional vectors which "support" the plane.

$$
\begin{cases}
x = 1 + s - t \\
2 - y = s \\
z + 1 = 2s
\end{cases}
$$

Composistion of the last two equations

$$
\begin{cases}
x = 1 + s - t \\
z + 1 = 2(2 - y)
\end{cases}
$$

$$
\begin{cases}
x = 1 + s - t \\
2y + z = 3
\end{cases}
$$

What should I do about $x = 1 + s - t$ ?

 

[andrewkirk]

You can use the second equation to eliminate $s$ from $x=1+s-t$. You will be left with an equation involving $x,y,t$. Since $t$ can be any real number, and there are no other equations that can enable us to eliminate $t$, what does that tell you about the dependence of the value of $x$ on $y$, or the lack of such dependence? What about $x$'s dependence or lack thereof on $z$?

If there is no dependence on either, what does that tell you about the plane? Is it parallel to any landmark of interest?

 

[ramzerimar]

Okay, you realize that you are dealing with parametric equations for your plane. Remember that, for determining the equation of a plane you just need two informations: a normal vector to the plane, and a vector parallel to the plane, and you should be able to find the equation by taking the dot product.

So, how do you determine a normal vector using those parametric equations?

 

[Rectifier]

You can use the second equation to eliminate $s$ from $x=1+s-t$.

Are you referring to
$$
\begin{cases}
x = 1 + s - t \\
2y + z = 3
\end{cases}
$$
?

If yes, how can I do that? Do i write 3 as 2 + 1, subtract 2 from both sides and then substitute the expression for 1 from both equations?

...a normal vector to the plane, and a vector parallel to the plane, and you should be able to find the equation by taking the dot product.

Unfortunately I am not familiar with that.

 

[andrewkirk]

Are you referring to

No, the second equation in the original set of three, which is $y=2-s$.

 

[Rectifier]

No, the second equation in the original set of three, which is $y=2-s$.

Alright, but even if I did that I would still have that top equation with the abundant parameter $t$ left.

 

[andrewkirk]

Yes, as I pointed out in post 2. Read over post 2, which suggests how you can use that fact to form conclusions.

 

[Rectifier]

What do you mean by dependence?

 

[andrewkirk]

In this context, by '$x$ depends on $y$' I mean that choosing a value of $y$ in some way narrows down the range of possible values for $x$. With those equations, does choosing a value for $y$ do that? Or can we choose $x$ to be any number at all for a given value of $y$?

 

[Rectifier]

I am not sure if I get the point in the post above but should I be thinking in terms of a linear equation now? Since there are no more t:s to eliminate that one from the equation, doesn't it matter what values x and y have in that equation then since we can't find a value for $t$?

Sorry if I missed your point.

 

[andrewkirk]

We have two equations:
$$2y+z=3$$
$$x=3-y-t$$
The first one tells us that for every value of $z$ there is a unique possible value of $y$. It does not constrain $x$ at all.
Does the second equation constrain $x$ at all, given that $t$ can be any real number? For example, if $y$ is 1, can you find a value of $t$ that makes $x$ equal to 3? What about a value of $t$ that makes $x=-3$, or $x=1,000,000$? Now what if $y=-10$, given any potential value for $x$, can you find a value of $t$ that makes $x$ equal to that value? If so, then the second equation does not constrain $x$ at all. Neither equation does. What set of points do you get if you apply the constraint of the equation $2y+z=3$ and let $x$ be anything at all?

 

[Rectifier]

We have two equations:
$$2y+z=3$$
$$x=3-y-t$$
The first one tells us that for every value of $z$ there is a unique possible value of $y$. It does not constrain $x$ at all.
Does the second equation constrain $x$ at all, given that $t$ can be any real number? For example, if $y$ is 1, can you find a value of $t$ that makes $x$ equal to 3? What about a value of $t$ that makes $x=-3$, or $x=1,000,000$? Now what if $y=-10$, given any potential value for $x$, can you find a value of $t$ that makes $x$ equal to that value? If so, then the second equation does not constrain $x$ at all. Neither equation does.

So for each $y$ there is a $t$ that makes the equation disappear ( $x = x$ or $0 = 0$ ) since $t$ can be any real number.

What set of points do you get if you apply the constraint of the equation $2y+z=3$ and let $x$ be anything at all?

That set of points is still a plane, right?

 

[andrewkirk]

Yes. How is the plane oriented?

 

[Rectifier]

Yes. How is the plane oriented?

It is parallel to the x - plane. I cheated a little since I checked the key in my book before I asked the question on here :D. But I guess you could say that it is parallel since there are no x - in the last surviving equation.

Thank you for your help.

 

[Mark44]

Here's a different way to look at things.

The problem
I am trying to write the equation for the plane on the following form $ax + by + cz + d = 0$

$x = 1 + s - t \\ y = 2 - s \\ z = -1 + 2s$

$\begin{bmatrix} x \\ y \\z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + s\begin{bmatrix} 1 \\ -1 \\ 2\end{bmatrix} + t\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$
Geometrically, an arbitrary point P(x, y, z) in the plane can be reached from the origin by the sum of a vector $\vec{OP_0}$ (with P₀(1, 2, -1) ), plus a linear combination of $\vec{u}$ and $\vec{v}$, the last two vectors shown in the equation above. Those two vectors lie in the plane, and being nonparallel, define the plane's orientation.

It's fairly straightforward to determine the standard equation of a plane, knowing two vectors that lie in the plane, and a point in the plane.

 

[ramzerimar]

Unfortunately I am not familiar with that.

Dot product.

Also, refer to @Mark44 method.