Equation of a plane parametric and cartesian

**terrorsquid** · October 15th 2011, 12:27 AM

I have three points $A = (2,0,1)$ , $B = (1,1,2)$ and $C = (0,0,3)$ that are the vertices of a triangle in $\mathbb{R}^3$ . I am trying to find the equation of the plane on which the triangle lies in parametric and cartesian form.

I have an answer written down:

$(x,y,z) = (2,0,1) + \lambda (-1,1,1) + \mu (-1,-1,1)$

But I'm not sure what method was used to achieve this answer. The way I did it is:

$\vec{AB} = (-1,1,1)$

$\vec{BC} = (-1,-1,1)$

$\vec{AB} \times \vec{BC} = 2x-0y+2z$

$\therefore (x,y,z) = 2(x-2)+2(z-1)$

$x+z = 3$ (cartesian)

Is this right and how do I get the parametric form?

**HallsofIvy** · October 15th 2011, 09:41 AM

Originally Posted by terrorsquid

I have three points $A = (2,0,1)$ , $B = (1,1,2)$ and $C = (0,0,3)$ that are the vertices of a triangle in $\mathbb{R}^3$ . I am trying to find the equation of the plane on which the triangle lies in parametric and cartesian form.

I have an answer written down:

$(x,y,z) = (2,0,1) + \lambda (-1,1,1) + \mu (-1,-1,1)$

But I'm not sure what method was used to achieve this answer. The way I did it is:

$\vec{AB} = (-1,1,1)$

$\vec{BC} = (-1,-1,1)$

$\vec{AB} \times \vec{BC} = 2x-0y+2z$

Well, no, the cross product is <2, 0, 2>. There are no "x", "y", or "z" in it.

$\therefore (x,y,z) = 2(x-2)+2(z-1)$

I don't know what you mean by "(x,y,z)= ". Given that <2, 0, 2> is a normal vector, the plane is of the form $2(x- x_0)+ 0(y- y_0)+ 2(z- z_0)= 0$ . If you choose (2, 0, 1) as your point, then 2(x- 2)+ 2(z- 1)= 0 which leads to 2x- 4+ 2z- 2= 0, 2x+ 2z= 6 and dividing by 2 gives your result.

$x+z = 3$ (cartesian)

Is this right and how do I get the parametric form?

It's easy to check it yourself, isn't it? (x,y,z)= (2, 0, 1) gives 2+ 1= 3, which is true. (1, 1, 2) gives 1+ 2= 3 which is true. (0, 0, 3) gives 0+ 3= 3, which is true. The three given points are on this plane and, geometrically, only one plane contains all of three independent points.

Another way you could have gotten this was to write Ax+ By+ Cz= D and put your given points into it: (2, 0, 1) gives 2A+ C= D. (1, 1, 2) gives A+ B+ 2C= D. (0, 0, 3) gives 3C= D. Multiplying the equation by a number does not change the plane so we can always choose any one constant and still have the same plane. In order to get an integer, take D= 3 so that C= 1. Now the first equation becomes 2A+ 1= 3 so A= 1. Finally, we have 1+ B+ 2= 3 and so B= 0. x+ z= 3 as you have.

As for getting the parametric equations, you can do this. z+ x= 3 means z= 3- x. y could be anything so we can let x and y be anything, say x= u, y= v, and then have z= 3- u. The parametric equations are x= u, y= v, z= 3- u.

A more direct way, without finding the Cartesian equation first, is to note that the vector from (2, 0, 1) to (1, 1, 2) is, as you say, (1- 2, 1- 0, 2- 1)= (-1, 1, 1), and the vector from (1, 1, 2) to (0, 0, 3) is (0- 1, 0- 1, 3- 2)= (-1, -1, 1). Since those vectors are independent, any vector in the plane can be written as a linear combination of then and so any point in the plane can be written in the form (x, y, z)= (2, 0, 1)+ s(-1, 1, 1)+ t(-1, -1, 1) which gives x= 2- s- t, y= s- t, z= 1+ s+ t.

You understand, I hope, that there are many choices for parametric equations of a given plane. These, x= u, y= v, z= 3- u, and x= 2- s- t, y= s- t, z= 1+ s+ t may look very different but note that z= 1+ s+ t= 3- (2- s- t)= 3- x. Taking u= 2- s- t= 2-(s+t) and v= s- t gives the same paratric equations as before.

**terrorsquid** · October 15th 2011, 06:14 PM

Originally Posted by HallsofIvy

Well, no, the cross product is <2, 0, 2>. There are no "x", "y", or "z" in it.

I add x, y and z to my matrix when finding the determinant; it's just a method I like which gives an equation and I just extract the coefficients.

I don't know what you mean by "(x,y,z)= ". Given that <2, 0, 2> is a normal vector, the plane is of the form $2(x- x_0)+ 0(y- y_0)+ 2(z- z_0)= 0$ . If you choose (2, 0, 1) as your point, then 2(x- 2)+ 2(z- 1)= 0 which leads to 2x- 4+ 2z- 2= 0, 2x+ 2z= 6 and dividing by 2 gives your result.

Yah, I wasn't sure how to write it; I thought it was just standard notation. Maybe $f(x,y,z)$ is more correct?

You understand, I hope, that there are many choices for parametric equations of a given plane. These, x= u, y= v, z= 3- u, and x= 2- s- t, y= s- t, z= 1+ s+ t may look very different but note that z= 1+ s+ t= 3- (2- s- t)= 3- x. Taking u= 2- s- t= 2-(s+t) and v= s- t gives the same paratric equations as before.

Yes, I was playing around with different choices for the vectors and points and any point + scalar multiples of any two vectors that went from one point to another on the plane would suffice - they all produced the same cartesian form if I was to solve the values for x, y and z simultaneously.

Thanks.

**Prove It** · October 15th 2011, 07:46 PM

Originally Posted by terrorsquid

I have three points $A = (2,0,1)$ , $B = (1,1,2)$ and $C = (0,0,3)$ that are the vertices of a triangle in $\mathbb{R}^3$ . I am trying to find the equation of the plane on which the triangle lies in parametric and cartesian form.

I have an answer written down:

$(x,y,z) = (2,0,1) + \lambda (-1,1,1) + \mu (-1,-1,1)$

But I'm not sure what method was used to achieve this answer. The way I did it is:

$\vec{AB} = (-1,1,1)$

$\vec{BC} = (-1,-1,1)$

$\vec{AB} \times \vec{BC} = 2x-0y+2z$

$\therefore (x,y,z) = 2(x-2)+2(z-1)$

$x+z = 3$ (cartesian)

Is this right and how do I get the parametric form?

Why go through all of this? Since you have three points, substitute each of them into $\displaystyle ax + by + cz = d$ to get three equations so solve simultaneously. Your answer will be in terms of $\displaystyle d$ , but then you can substitute one of the three points to solve for $\displaystyle d$ .

**terrorsquid** · October 15th 2011, 08:25 PM

Originally Posted by Prove It

Why go through all of this? Since you have three points, substitute each of them into $\displaystyle ax + by + cz = d$ to get three equations so solve simultaneously. Your answer will be in terms of $\displaystyle d$ , but then you can substitute one of the three points to solve for $\displaystyle d$ .

So, I would have:

$2x+z=d$
$x+y+2z=d$
$3z=d$

I don't understand.

**Prove It** · October 15th 2011, 08:35 PM

Originally Posted by terrorsquid

So, I would have:

$2x+z=d$
$x+y+2z=d$
$3z=d$

I don't understand.

So from the third equation $\displaystyle z = \frac{d}{3}$ .

Substitute into the first equation to get $\displaystyle 2x + \frac{d}{3} = d \implies 2x = \frac{2d}{3} \implies x = \frac{d}{3}$ .

Substitute into the second equation to get $\displaystyle \frac{d}{3} + y + \frac{2d}{3} = d \implies y = 0$ .

So the equation of your plane is

$\displaystyle \frac{d}{3}x + 0y + \frac{d}{3}z = d$ .

Now substitute one of the three points to evaluate d.

Edit: Never mind, we need a fourth point to evaluate d.

**terrorsquid** · October 15th 2011, 09:03 PM

Since they are not parallel, they must intersect somewhere. What is the best way to find their intersection? The method I have seen goes as follows:

$4x-y+2z=5$
$7x-3y+4z=8$

eliminate z...

$x+y=2$

$y=2-x$

eliminate y...

$5x+2z=7$

$z=\frac{7}{2}-\frac{5x}{2}$

then let $x=t$ :

$x=t$

$y=2-t$

$z=\frac{7}{2}-\frac{5t}{2}$

I'm a little confused by this though; how do I derive a Cartesian form from it?

**Deveno** · October 15th 2011, 09:17 PM

Originally Posted by Prove It

So from the third equation $\displaystyle z = \frac{d}{3}$ .

Substitute into the first equation to get $\displaystyle 2x + \frac{d}{3} = d \implies 2x = \frac{2d}{3} \implies x = \frac{d}{3}$ .

Substitute into the second equation to get $\displaystyle \frac{d}{3} + y + \frac{2d}{3} = d \implies y = 0$ .

So the equation of your plane is

$\displaystyle \frac{d}{3}x + 0y + \frac{d}{3}z = d$ .

Now substitute one of the three points to evaluate d.

Edit: Never mind, we need a fourth point to evaluate d.

not really. your final equation is d/3(x + z) = d, that is d(x + z) = 3d, which implies x+z = 3

(the case d = 0 leads to the entire xz-plane, but we know we have points with non-zero y-coordinate on our plane).

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