Finding the Equation of a Plane with Given Intercepts

[Lancelot59]

Hello everyone. This problem asks me to find the equation of a plane with an x-intercept of a, a y-intercept of b, and a z-intercept of c.

I started off with the scalar plane equation:

\hat{i}(x-x_{o})+\hat{j}(y-y_{o})+\hat{k}(z-z_{o})+d=0

I decided to start with Y, stating that when x=z=0 y=b. I then got this after moving some stuff around:

-\hat{i}(x_{o})+\hat{j}(b-y_{o})-\hat{k}(z_{o})+d=0
\hat{j}(b-y_{o})=\hat{i}(x_{o})++\hat{k}(z_{o})+d

It was at this point I ran out of ideas. I still have too many unknowns, and I can see solving for the others just making the equation cancel out.

How can I go about solving this problem?

 

[Rasalhague]

How could you use those three given points to find a vector at right angles to the plane?

 

[Lancelot59]

a=(x,0,0)
b=(0,y,0)
c=(0,0,z)

I'm not entirely sure. So I could take vectors between those coordinates...but would those not end up being parallel with the plane? Unless I take a cross product between them. I'll give it a try.

 

[Mark44]

No, the three points are (a, 0, 0), (0, b, 0), and (0, 0, c). Call these points A, B, and C. The vectors AB and AC lie in the plane (as does BC). AB X AC gives you a vector that is perpendicular to the plane.

 

[Lancelot59]

I see. Thanks for the help!