# Planes in 3-space; Equations of planes

• Mar 29th 2008, 06:46 AM
mathsToday
Planes in 3-space; Equations of planes
Hi guys,

I've got this exercise where I'm required to describe the equation of the plane determined by the fact that the following
three points lie in it: (−7, 1, 0), (2, −1, 3), (4, 1, 6).

What i've done for it is as followings:

To describe the equation of the plane we require a point in the plane and a vector $n$ that is orthogonal to every vector in the plane. The vectorn is said to be normal to the plane.

We suppose our point in the plane to be $P(x0,y0, z0)$
We chose n the normal vector to be on of our given three points.
$n = (4, 1, 6)$
Now we have discovered a point within the plane and a normal we can carry out the equation of the form

$a(x-x0)+b(y-y0)+c(z-z0)$
where $a,b,c = 4, 1, 6$

Well thats my attempt at the question i not quiet sure that i've answered it fully but i think i did most part correctly. I would appreciate any advice/help with it with the question it self and my answer (if it could be built on).

Thanks a lot !
• Mar 29th 2008, 07:13 AM
Plato
You have made some mistake in calculations.
Did you use the cross product?
I get $a=12,\;b=21,\;\&\,c=-22.$
• Mar 29th 2008, 07:28 AM
mathsToday
Quote:

Originally Posted by Plato
You have made some mistake in calculations.
Did you use the cross product?
I get $a=12,\;b=21,\;\&\,c=-22.$

Mistakes in which area or as whole ?
erm...no i've not used cross product i wasn't aware i needed it to answer the question. So i should be using it and how?

• Mar 29th 2008, 08:29 AM
TheEmptySet
Quote:

Originally Posted by mathsToday
Mistakes in which area or as whole ?
erm...no i've not used cross product i wasn't aware i needed it to answer the question. So i should be using it and how?

You need to find two vectors that lie in the plane

$v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k$

and

$v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-11 && 0 && -6 \\
-2 && -2 && -3 \\
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k
$

This is the normal vector to the plane

Good luck
• Mar 29th 2008, 08:50 AM
mathsToday
Quote:

Originally Posted by TheEmptySet
You need to find two vectors that lie in the plane

$v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k$

and

$v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-11 && 0 && -6 \\
-2 && -2 && -3 \\
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k
$

This is the normal vector to the plane

Good luck

thanks a lot

Oh ok it wasn't explained to me this way(Worried). And then so i use the found normal in the equation i proposed early ?
• Mar 29th 2008, 01:45 PM
mathsToday
ahh i get it a lot better now actually(Yes)..think i may have answered it

Thanks for help guys (Talking)
• Apr 2nd 2008, 04:45 AM
mathsToday
Quote:

Originally Posted by TheEmptySet
You need to find two vectors that lie in the plane

$v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k$

and

$v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-11 && 0 && -6 \\
-2 && -2 && -3 \\
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k
$

This is the normal vector to the plane

Good luck

We the cross product i get $12i + 21k + 22k$ am i correct or have I carried out the wrong arithmetic?.
• Apr 2nd 2008, 10:18 AM
TheEmptySet
$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-11 && 0 && -6 \\
-2 && -2 && -3 \\
\end{vmatrix}
$

Expanding along the first row we get...

$i\cdot \begin{vmatrix}
0 && -6 \\
-2 && -3 \\
\end{vmatrix} -j \cdot \begin{vmatrix}
-11 && -6 \\
-2 && -3 \\
\end{vmatrix}+ k \cdot \begin{vmatrix}
-11 && 0 \\
-2 && -2 \\
\end{vmatrix}=$

$(0(-3)-(-2)(-6)) \vec i -((-11)(-3)-(-2)(-6)) \vec j +((-11)(-2)-(-2)(0)) \vec k$

$-12 \vec i -21 \vec j + 22 \vec k \iff 12 \vec i + 21 \vec j -22 \vec k$

Both of these are normal to the plane.
• Apr 6th 2008, 05:47 AM
mathsToday
Quote:

Originally Posted by TheEmptySet
You need to find two vectors that lie in the plane

$v_1= \underbrace{(-7,1,0)}_{Head}-\underbrace{(4,1,6)}_{tail}=-11 \vec i -6 \vec k$

and

$v_2= \underbrace{(2,-1,3)}_{Head}-\underbrace{(4,1,6)}_{tail}=-2 \vec i -2 \vec j -3 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-11 && 0 && -6 \\
-2 && -2 && -3 \\
\end{vmatrix} = 12 \vec i +21 \vec j- 22 \vec k
$

This is the normal vector to the plane

Good luck

Furthering the equation i get

$12(x-x0) + 21(y-y0) -22(z-z0) = 0$

where as shown i obtained 12, 21 and -22 by calculating the normal. I then substitute a vector in the plane, i.e 4,1,6), giving you an equation

$12(x-4) + 21(y-1) -22(z-6) = 0.$

The equation can now be rewritten to the form
$ax+by+cz = d$

then
$12x+21y+-22z = d$

I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

Once this is done then I should be able to check my equation by seeing whether each of the given three points in the plane satisfies it right ?:)

Thanks a bunch
• Apr 6th 2008, 06:40 AM
TheEmptySet
Quote:

Originally Posted by mathsToday
Furthering the equation i get

$12(x-x0) + 21(y-y0) -22(z-z0) = 0$

where as shown i obtained 12, 21 and -22 by calculating the normal. I then substitute a vector in the plane, i.e 4,1,6), giving you an equation

$12(x-4) + 21(y-1) -22(z-6) = 0.$

The equation can now be rewritten to the form
$ax+by+cz = d$

then
$12x+21y+-22z = d$

I'm not sure how to reorganize that equation to evaluate the d mentioned above. Any help /push in the right direction would be great!

Once this is done then I should be able to check my equation by seeing whether each of the given three points in the plane satisfies it right ?:)

Thanks a bunch

$12(x-4) + 21(y-1) -22(z-6) = 0$

Distribute out the left hand side.

$12x-48 + 21y-21 -22z+132 = 0 \iff 12x +21y-22z +63=0$

so

$12x+21y-22z=-63$

P.S. if you want a sub script (underscore) use the following code

$
12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0$

12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0
• Apr 6th 2008, 07:23 AM
mathsToday
Quote:

Originally Posted by TheEmptySet
$12(x-4) + 21(y-1) -22(z-6) = 0$

Distribute out the left hand side.

$12x-48 + 21y-21 -22z+132 = 0 \iff 12x +21y-22z +63=0$

so

$12x+21y-22z=-63$

P.S. if you want a sub script (underscore) use the following code

$
12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0$

12(x-x_0) + 21(y-y_0) -22(z-z_0) = 0

ahhh i see now.

Thanks so much, your help been amazing useful, learnt what I had no clue about in class:)
• Jul 20th 2008, 10:28 AM
mathsToday
So i've got a new equation that needs to be determined by the the fact the following three points lie in it: $(1,2,-4), (2,3,7)$ and $(4,-1,3)$ So i find two vectors that lie in the plane.

$v_1= \underbrace{(1,2,-4)}_{Head}-\underbrace{(4,-1,3)}_{tail}= -3 \vec i 3 \vec j -7 \vec k$

and

$v_2= \underbrace{(2,3,7)}_{Head}-\underbrace{(4,-1,3)}_{tail}=-2 \vec i -4 \vec j -4 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-3 && 3 && -7 \\
-2 && -4 && -4 \\
\end{vmatrix} = 16 \vec i +2 \vec j- 18 \vec k
$

This is the normal vector to the plane

I just want to check that what i've done is correct so far?
• Jul 20th 2008, 11:27 AM
Plato
The ideas are correct.
BUT, you have made many sign errors in almost all the above circulations.
Go back and check each one again.
• Jul 30th 2008, 01:50 PM
mathsToday
$v_1= \underbrace{(1,2,-4)}_{Head}-\underbrace{(4,-1,3)}_{tail}= -3 \vec i, 3 \vec j, -7 \vec k$

and

$v_2= \underbrace{(2,3,7)}_{Head}-\underbrace{(4,-1,3)}_{tail}=-2 \vec i, 4 \vec j , 4 \vec k$

To find a normal vector we need to take the cross product

$\begin{vmatrix}
\vec i && \vec j && \vec k \\
-3 && 3 && -7 \\
-2 && 4 && 4 \\
\end{vmatrix} = 40 \vec i +26 \vec j- 6 \vec k
$

This is the normal vector to the plane

I just want to check that my changes are now correct? thanks.
• Jul 30th 2008, 04:01 PM
mathsToday
Quote:

Originally Posted by TheEmptySet
$12(x-4) + 21(y-1) -22(z-6) = 0$

Distribute out the left hand side.

$12x-48 + 21y-21 -22z+132 = 0 \iff 12x +21y-22z +63=0$

so

$12x+21y-22z=-63$

I really can't remember how we find the answer -63 and can't seem to work it out either (Worried) can anybody run through the steps to getting the resulting -63?? thanks a lot!