# Homework Help: Equation of plane

1. Nov 10, 2012

### Miike012

The question and solution is in the paint doc. My concern was how did they find the equation of the plane without finding the vector normal to the plane? Im guessing they found the vector but left out the steps...

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2. Nov 11, 2012

### ehild

They could find the normal vector, and got the equation of the plane using it.

ehild

Last edited: Nov 11, 2012
3. Nov 11, 2012

### SammyS

Staff Emeritus
The equation for a plane can be given in the form, $ax+by+cz=d\ .$ Of course, one of those constants is arbitrary.

Just plug-in the each set of coordinates for the three intercepts, individually, to find $\displaystyle \frac{a}{d}\,,\ \frac{b}{d}\,\ \text{ and },\ \frac{c}{d}\,,\$ then find a convenient value to use for d.

4. Nov 11, 2012

### ehild

That is much easier...

ehild

5. Nov 11, 2012

### Miike012

wouldn't it be a = d/x, b = d/y, and c = d/z where (x,y,z) are points on the plane and <a,b,c> is the normal vector
hence if (x,y,z) = (0,0,z) then cz = d and c = d/z .... you can do the same for a and b components of the normal.

6. Nov 11, 2012

### SammyS

Staff Emeritus
For the x-intercept, x=1, y=0, and z=0.

Therefore, a(1) + 0 + 0 = d  →  a/d =1,  etc.