Plane equation understanding help

Jan 2017
9
0
Houston
I'm given three points A(1,-2,1), B(3,2,2) and C(-2,1,-5) and I'm asked to find plane equation for it. I know to start it out, you'd take dot product of one of components with others, then take the cross product of that to get your x0, y0, and z0 but after that I'm confused. Would someone mind helping me out to understand how the answer gets to 3x-y-2z-3=0? Thanks in advance.

Dot products: AB=(2,4,1) and AC=(-3,3,-6)
Cross product: ABxAC=(-27,9,18)
 
Last edited:

Plato

MHF Helper
Aug 2006
22,473
8,643
I'm given three points A(1,-2,1), B(3,2,2) and C(-2,1,-5) and I'm asked to find plane equation for it. I know to start it out, you'd take dot product of one of components with others, then take the cross product of that to get your x0, y0, and z0 but after that I'm confused. Would someone mind helping me out to understand how the answer gets to 3x-y-2z-3=0? Thanks in advance.

Dot products: AB=(2,4,1) and AC=(-3,3,-6)
Cross product: ABxAC=(-27,9,18)
$N = \overrightarrow {AB} \times \overrightarrow {AC} ,\quad P = \left\langle {x - {a_1},y - {a_2},z - {a_3}} \right\rangle $

The plane is $N\cdot P=0$
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I'm given three points A(1,-2,1), B(3,2,2) and C(-2,1,-5) and I'm asked to find plane equation for it. I know to start it out, you'd take dot product of one of components with others, then take the cross product of that to get your x0, y0, and z0 but after that I'm confused.
No, you are confused well before that! The dot product of two vectors is a number not a vector so you can't take the cross product with anything! Also it makes no sense to talk about the dot product of "components". Vectors have "components" but you take a dot product of vectors, not their components. Here, you are given three points. Points have "coordinates", not "components". Subtract the coordinates of, say, A from the coordinates of the two other points, B, and C to get two vectors \(\displaystyle \vec{BA}\) and \(\displaystyle \vec{CA}\) that lie in the plane. The cross product of those two vectors is perpendicular to both vector so perpendicular to the plane. The plane with perpendicular vector <P, Q, R> containing point \(\displaystyle (x_0, y_0, z_0)\) has equation \(\displaystyle P(x- x_0)+ Q(y- y_0)+ R(z- z_0)= 0\)

quote] Would someone mind helping me out to understand how the answer gets to 3x-y-2z-3=0? Thanks in advance.

Dot products: AB=(2,4,1) and AC=(-3,3,-6)
Cross product: ABxAC=(-27,9,18)[/QUOTE]
 
Jan 2017
9
0
Houston
Holy , I'm dumb. Didn't even realize that... Well, I wasted a lot of time getting frustrated why I couldn't figure it out.. lol