# Construct A Plane From A Coordinate And A Unit Vector And Find Its X,Y,Z Angles

#### SESruss

I'm trying to learn how to figure out the X,Y,Z rotation angles of a plane so that it is normal to a unit vector. I have a coordinate point that is on the plane and a unit vector that is normal to plane to start with. Appreciate any help I can get.

#### HallsofIvy

MHF Helper
Let (a, b, c) be the point and <P, Q, R> the normal vector. Then any point (x, y, z) that lies on that plane satisfies <x- a, y- b, z- c>.<P, Q, R>= P(x- a)+ Q(y- b)+ R(z- c)= 0. It is easy to get the angles from the equation of the plane.

#### SESruss

So for a point(0,0,0) and a unit vector (-.0088, .0030, -.9999) the equation would be:

-.0088(x-0) + .003(y-0) -.9999(z-0) = 0
-.0088x + .003y -.9999z = 0

In order to find the angles of the plane I tried calculating the angle between two planes. My thought was the axes unit vectors would be:

(1,0,0) for x
(0,1,0) for y
(0,0,1) for z

And their equations of a plane with the same point (0,0,0):

x=0
y=0
z=0

Using the equation:

cos α = |A1·A2 + B1·B2 + C1·C2|
√A1² + B1² + C1² √A2² + B2² + C2²

The results seem wrong so I believe I'm on the wrong track.

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