converting parametric to cartesian equation

[roam]

1. I found the parametric equation of a plane;

\left(\begin{array}{ccc}x\\y\\z\end{ar ray}\right) = \left(\begin{array}{ccc}1\\2\\3\end{ar ray}\right) +s\left(\begin{array}{ccc}1\\1\\0\end{ar ray}\right) +t \left(\begin{array}{ccc}2\\1\\-1\end{ar ray}\right)

s,t ∈ R.

I was asked to find a Cartesian equation. So I write down the three equations;

x=1+s+2t
y=2+s+t
z=3−t

I don't understand how this set can be solved. What's the aim? Do I need to eliminate s,t from the equations?

3. The attempt at a solution

1) x=1+s+2t
2) y=2+s+t
3) z=3−t

If I subtract (1) and (2) => x–y=t–1
Subtracting the third from the second => y-z = s-1+2t
And I can also find an expression for t, from the (3) => t=3-z

Could you please show me how we can use this info to find the cartesian equation of the plane.

[Office_Shredder]

You have t=3-z. This is an excellent start, as you can go back to say x-y=t-1 and exchange t with 3-z. Then you have an equation in x,y and z

[roam]

So as soon as we have an equation in x,y and z it will be the cartesian equation?
So, therefore x–y+z=2 is the cartesian equation of this plane.

[Mark44]

So as soon as we have an equation in x,y and z it will be the cartesian equation?
So, therefore x–y+z=2 is the cartesian equation of this plane.

That will work. A plane can also be defined by a point in the plane and a vector that is normal to the plane. In your case, the point (1, 2, 3) is in the plane. To get a normal, cross the two other vectors in your parametric equation, (1, 1, 0) and (2, 1, -1). The cross product will give you a vector (A, B, C). The equation of the plane will be A(x - 1) + B(y - 2) + C(z - 3) = 0.

What I've described is probably a bit more complicated, but has the advantage of helping you comprehend the plane geometrically.

[roam]

Oh, thanks very much!
Yes, I know how to do it by finding the normal using the cross product, I just wasn't sure how to solve those set of equations.

[roam]

Just out of curiosity, I tried the method you suggested:

direction vectors of the plane (vectors parallel to the plane) are:

u = \left(\begin{array}{ccc}1\\1\\0\end{ar ray}\right) and v = \left(\begin{array}{ccc}2\\1\\-1\end{ar ray}\right)

so that a normal vector is u ×v (cross product) = \left(\begin{array}{ccc}-1\\-1\\-1\end{ar ray}\right)

So the cartesian equation is -x - 13y −z = d, satisfying any point on the plane. For example if I substitute a point on the plane (1,2,3), [which is the position vector];

-1-2-3 = d
d = -6

=> -x - 13y −z = -6

But this is different from what we got previously, x–y+z=2. Why is it different?
Thanks again!

[HallsofIvy]

Just out of curiosity, I tried the method you suggested:

direction vectors of the plane (vectors parallel to the plane) are:

u = \left(\begin{array}{ccc}1\\1\\0\end{ar ray}\right) and v = \left(\begin{array}{ccc}2\\1\\-1\end{ar ray}\right)

so that a normal vector is u ×v (cross product) = \left(\begin{array}{ccc}-1\\-1\\-1\end{ar ray}\right)
??Certainly not! Where in the world would you get the "13" from? The cross product Of \vec{i}+ \vect{j} and 2\vec{i}+ \vec{j}- \vec{k} is -\vec{i}+ \vec{j}- \vec{k} or \vec{i}- \vec{j}+ \vec{k}} depending on the order of multiplication.

So the cartesian equation is -x - 13y −z = d, satisfying any point on the plane. For example if I substitute a point on the plane (1,2,3), [which is the position vector];

-1-2-3 = d
d = -6

=> -x - 13y −z = -6

But this is different from what we got previously, x–y+z=2. Why is it different?
Thanks again![/QUOTE]

[roam]

??Certainly not! Where in the world would you get the "13" from?

oops there was a typo in my previous post...so I'll start over.

I want the general cartesian equation of the plane, Ax+By+Cz = D

I found the cross product of the two direction vectors;

\left(\begin{array}{ccc}i&j&k\\1&1&0\\2&1&-1\end{ar ray}\right)

i \left(\begin{array}{ccc}1&0\\1&-1\end{ar ray}\right), -j \left(\begin{array}{ccc}1&0\\2&-1\end{ar ray}\right), k \left(\begin{array}{ccc}1&1\\2&1\end{ar ray}\right)

Cross Product = (-1,-1,-1)

The cartesian equation is -x-y-z = D

To find a D I just substitute point (1,2,3)
-1-2-3 = -6

Therefore => -x-y-z = -6

But the problem is that this is different from the equation we obtained previously, x–y+z=2. Why is that? I don't understand it.

[HallsofIvy]

oops there was a typo in my previous post...so I'll start over.

I want the general cartesian equation of the plane, Ax+By+Cz = D

I found the cross product of the two direction vectors;

\left(\begin{array}{ccc}i&j&k\\1&1&0\\2&1&-1\end{ar ray}\right)

i \left(\begin{array}{ccc}1&0\\1&-1\end{ar ray}\right), -j \left(\begin{array}{ccc}1&0\\2&-1\end{ar ray}\right), k \left(\begin{array}{ccc}1&1\\2&1\end{ar ray}\right)

Cross Product = (-1,-1,-1)
No, the "j" determinant is not 1, so the "j" component is not -1.

The cartesian equation is -x-y-z = D

To find a D I just substitute point (1,2,3)
-1-2-3 = -6

Therefore => -x-y-z = -6

But the problem is that this is different from the equation we obtained previously, x–y+z=2. Why is that? I don't understand it.

[roam]

I understand. It's (-1,1,-1)

=> -x+y-z = -2

Why is it different from the other equation we obtained (y+z=2)? It is still a multiple of that equation. It looks like it has been multiplied through by -1.

[Mark44]

-x + y -z = -2 is different from x + y + z = 6. Geometrically they represent different planes in R^3.

[roam]

I know! I'm talking about -x+y-z = -2 VS x–y+z=2 ?