# Thread: Planes in 3-space; Describing the equations of a plane

1. ## Planes in 3-space; Describing the equations of a plane

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: $\displaystyle 1, 2, -4), (2, 3, 7), (4, -1, 3)$

So we find two vectors that lie in the plane

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

and

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

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

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

$\displaystyle 40(x-4) + 26(y-1) -6(z-3) = 0$

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

$\displaystyle 40x+26y-6z = d$

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!

So we distribute out the left hand side.

$\displaystyle 40x-160 + 26y-26 -6z+18 = 0 ?$

If any mistakes can be pointed out and any help towards the completion of this exercise would be greatly appreciated.

2. Note that we can also use $\displaystyle 20i + 13j - 3k$ as the normal.
Then the plane equation is $\displaystyle 20x + 13y - 3z = d$.
We can find $\displaystyle d$ simply by using one of the points.
Check it with another one of the points.

3. Originally Posted by Plato
Note that we can also use $\displaystyle 20i + 13j - 3k$ as the normal.
Then the plane equation is $\displaystyle 20x + 13y - 3z = d$.
We can find $\displaystyle d$ simply by using one of the points.
Check it with another one of the points.
Ok but the normal i've used is also correct isn't it?, along with my distrubiation of the LHS in the equation:
$\displaystyle 40(x-x0) + 26(y-y0) -6(z-z0) = 0$
"We can find simply by using one of the points. Check it with another one of the points."
That i'm not sure of, how do you mean check it?.

4. Originally Posted by mathsToday
Ok but the normal i've used is also correct isn't it?
The two vectors are parallel. The point is that it is simpler to use small numbers.

Originally Posted by mathsToday
"We can find simply by using one of the points. Check it with another one of the points." That i'm not sure of, how do you mean check it?.
Well use the point $\displaystyle (1,2, - 4):\;20(1) + 13(2) - 3( - 4) = 58$. So $\displaystyle d=58$.
Check with this point: $\displaystyle (4, - 1,3):\;20(4) + 13( - 1) - 3(3) = 58$.