Thread: Find an equation of the plane..

1. Find an equation of the plane..

Find an equation of the plane through the origin and parallel to the plane 3x + 2y + 1z = 4.

Since they are parallel, I tried to find the normal to 3x + 2y + 1z = 4 but i forget how.

Then use N . <x,y,z> = 0 (cause it passes through origin)

Is this correct, and if so, can i get help finding N?

Thanks

2. Re: Find an equation of the plane..

If Q= (1,0,1) and P=(2,1,-4) both points in the plane.

N . PQ = (n1,n2,n3) . (1,1,-5) = 0

N = (5,5,2). is that correct?

3. Re: Find an equation of the plane..

so equation of plane would be 5x + 5y + 2z = 0

4. Re: Find an equation of the plane..

Originally Posted by linalg123
Find an equation of the plane through the origin and parallel to the plane 3x + 2y + 1z = 4.
Any plane through the origin looks like $\displaystyle Ax+By+Cz=0~.$

Parallel planes have the same normals: $\displaystyle 3x+4y+z=~?$

5. Re: Find an equation of the plane..

Originally Posted by Plato
Any plane through the origin looks like $\displaystyle Ax+By+Cz=0~.$

Parallel planes have the same normals: $\displaystyle 3x+4y+z=~?$
Sorry, Where did you get 3x + 4y + z from?

6. Re: Find an equation of the plane..

Shortest approach would use the fact that the equations of the parallel planes have special property. Their coefficients on the left-hand side are proportional to each other.

Example: 3x + 2y + z = 4 is parallel to 6x+4y+2z=5. The second equation is equivalent to 3x+2y+z=5/2.
Example: 6x + 4y + 2z = 8 is the same plane as 3x + 2y + z = 4.

Although it's cumbersome, it's very useful for developing geometric intuition, to define the above property via normal vectors. The equation of a plane with point and normal vector looks like that:
$\displaystyle \vec{n} (\vec{x}-\vec{p}) = \vec{0}$ Where the vector x is the unknown which represents all points on the plane and the vector p represents the point on the plane we are given. (You may think of the coordinates of the points as the vectors from origin to the point.)
So we rework the equation to get $\displaystyle \vec{n}\vec{x}=\vec{n}\vec{p}$. When we expand the dot products it appears that the coefficients in front of our variables x1 (which we may as well rename to x), x2(which we may rename to y), and x3 (... to z) are just the components of the normal vector of the plane. So saying that "all parallel planes have the same normal vectors" is equivalent to saying "The equations of parallel planes may be reworked in such way that they will have the same left hand sides".

Using that knowledge to solve the task we need equation that has the form 3x+2y+z=d such that the origin (Point (0,0,0)) is a solution. This means that 3(0) + 2(0) + (0) = d must be true, which is only possible when d=0;

So our equation is 3x+2y+z=0.