Cartesian equation of the plane through the given points

[ezsmith]

Homework Statement

For each part, find the cartesian equation of the plane through the given points.
(1,0,3), (2,-4,3),(4,-1,2)

The Attempt at a Solution

No attempt. Dunno how to do :(

 

[HallsofIvy]

Haven't been paying attention in class? Use two pairs of those points to get two vectors in the plane: the vector from $(x_0, y_0, z_0)$ to $(x_1, y_1, z_1)$ is $(x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}$.

The cross product of two vectors is perpendicular to the two vectors so perpendicular to the plane they lie in.

The equation of a plane with normal vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ containing point $(x_0, y_0, z_0)$ is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

 

[ezsmith]

Haven't been paying attention in class? Use two pairs of those points to get two vectors in the plane: the vector from $(x_0, y_0, z_0)$ to $(x_1, y_1, z_1)$ is $(x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}$.

The cross product of two vectors is perpendicular to the two vectors so perpendicular to the plane they lie in.

The equation of a plane with normal vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ containing point $(x_0, y_0, z_0)$ is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.

I did listen in class but the lecturer only gave notes to copy during class and there is no example given in the notes so that is why. Anyway, I managed to solved it. Thanks a lot sir.. Really appreciate it :)