How Do You Find the Unit Normal to the Plane x + 2y - 2z = 15?

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Supra
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Homework Statement


Find the unit normal to the plane a + 2s - 2t = 15. What is the distance of the plane from the origin?

Homework Equations


The normal to a plane is given by s x t
For any plane, r.n = p [n = unit vector and p = constant]

The Attempt at a Solution


Not entirely sure what I'm meant to be doing here as I'm not given any real values for the vectors s and t, so I can't see how to crossing them achieves anything. I'd be capable of crossing the two vectors if they were in component form, but not here. Obviously I'm missing something here, any help would be great.

Many thanks,
/Supra.
 
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If a, s and t are vectors, that equation doesn't even make sense. The left side is a vector and the right side is a scalar. Are a, s and t the names of your coordinates?
 
My apologies, on reading the question again it seems the letters in the equation aren't meant to be vectors. So the equation is just a + 2s - 2t = 15 or to make it more simple: x + 2y - 2z = 15 where the letters are scalars I assume.
 
Supra said:
My apologies, on reading the question again it seems the letters in the equation aren't meant to be vectors. So the equation is just a + 2s - 2t = 15 or to make it more simple: x + 2y - 2z = 15 where the letters are scalars I assume.

That makes it easy, right? Now it's just your usual normal to a plane problem.