Question

What is the equation of the plane containing the triangle shown?

Answer 1

You are right!

Answer 2

Really? I accidentally clicked on that option, actually. I really could use an explanation on how to solve it. :)

Answer 3

@AmandaSmith69
Have you done vectors?

Answer 4

vectors.. I'm not sure actually. ._. math is not my strong point

Answer 5

From the figure, you can see that the plane passes through three points along the axes. These points are:
A(1,0,0), B(0,2,0), C(0,0,3).
From these three points, you can create two vectors that lie on the given plane, for example, P=BA=<1,-2,0>, and Q=CA=<1,0,-3>.
The normal vector of the plane can be found by the cross product PxQ, or

Since the plane passes through point (1,0,0), the equation of the plane is
6(x-1)+3(y-0)+2(z-0)=0
when simplified, gives
6x+3y+2z=6

(By the way, the response 2x+6y+3z=6 is incorrect).

A similar example is explained in detail (example 1) at the link below:
http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

and notes on cross product:
http://tutorial.math.lamar.edu/Classes/CalcII/CrossProduct.aspx

Answer 6

@mathmate I'm really confused about that entire thing, I'm sorry. I've been following geometry books and they are telling me to do something completely different. Eitherway, I'm still confused.