Equation of Plane: Solution + Finding Vector

[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? I am guessing they found the vector but left out the steps...

 

[ehild]

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

ehild

 

[SammyS]

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? I am guessing they found the vector but left out the steps...

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.

 

[ehild]

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.

That is much easier...

ehild

 

[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.

 

[SammyS]

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.

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

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