SOLUTION: Determine whether the following two planes x+4y-z = 7 and 5x-3y-7z = 11 are parallel, orthogonal, coincident (that is, the same) or none of these.

A normal vector to the plane x+4y-z = 7 is < 1,4,-1 >

A normal vector to the plane 5x-3y-7z = 11 is < 5,-3,-7 > 

Those are not proportional so the planes are not parallel.

The cosine of the angle between them is given by the dot product 
of the normals to the planes divided by the product of their 
magnitudes.

The dot product of their normals: 

< 1,4,-1 > • < 5,-3,-7 > = (1)(5)+(4)(-3)+(-1)(-7) = 5-12+7 = 0  

Aha. It's 0. So we don't need to divide by the product of their 
magnitudes.  The cosine of the angle between them is 0, so the
angle between them has cosine 0, which means the angle between 
them is 90°, so they are orthogonal.

Edwin