# Homework Help: Linear algebra - linear equation for a plane

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1. Oct 14, 2016

### 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 $x = 1 + s - t$ ?

Last edited: Oct 14, 2016
2. Oct 14, 2016

### 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?

3. Oct 14, 2016

### 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?

4. Oct 14, 2016

### Rectifier

Are you refering 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?

Unfortunately I am not familiar with that.

5. Oct 14, 2016

### andrewkirk

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

6. Oct 14, 2016

### Rectifier

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

7. Oct 14, 2016

### andrewkirk

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

8. Oct 14, 2016

### Rectifier

What do you mean by dependence?

9. Oct 14, 2016

### 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 $y$?

10. Oct 14, 2016

### 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.

11. Oct 14, 2016

### 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?

12. Oct 14, 2016

### Rectifier

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.

That set of points is still a plane, right?

13. Oct 14, 2016

### andrewkirk

Yes. How is the plane oriented?

14. Oct 14, 2016

### Rectifier

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.

Last edited: Oct 14, 2016
15. Oct 14, 2016

### Staff: Mentor

Here's a different way to look at things.
$\begin{bmatrix} x \\ y \\z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + s\begin{bmatrix} 1 \\ -1 \\ 2\end{bmatrix} + t\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$
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.

16. Oct 15, 2016

### ramzerimar

Dot product.

Also, refer to @Mark44 method.