# given a plane through origin and it's angle with X axis

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how to get the normal vector of the plane? the sample is frustum below: D3DXMATRIX *D3DXMatrixPerspectiveFovLH( D3DXMATRIX* pOut, // returns projection matrix FLOAT fovY, // vertical field of view angle in radians FLOAT Aspect, // aspect ratio = width / height FLOAT zn, // distance to near plane FLOAT zf // distance to far plane ); I know fovY and then how to get normal of a plane? and if I now fovY and I will get fovX by fovY*Aspect,Am I right?

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Well, to get a vector that is normal to the plane you must know the plane equation or to know two different vectors contained in the plane. That is:

1) Plane equation: (x-xo)*nx + (y-yo)*ny + (z-zo)*nz = 0

where the normal vector is N = (nx,ny,nz) and (xo,yo,zo) is a known point in the plane.

2) V1 and V2 are in the plane. So the normal vector can be defined by:

N = cross(V1,V2)

I didn't only understand what D3DXMatrixPerspectiveFovLH() has to do with that...

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I want to calculate the six plane of frustum, near plane ,far plane ,left plane ,right plane ,top plane ,bottom plane.

given a fovY ,the angle between top plane and bottom plane ,to calcuate the plane equation.
(these plane in view space so these through origin)
but get the plane equation ,I just need to know the normal of the plane.

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or say this way.
If I know a vector ,and rotate it a few angle aoubt some axis/x y or z/
so the new vector is ___?I think matrix multiplication work,but maybe exist
good simple method.

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