Hi,
I've been asked:
What does the plane 2x + 3y -z = 1 make with the x-y plane.
I'm completely stuck, never been any good with vectors, I'm sure the solution is simple using cos theta = n1 x n2/ modulus n1 x modulus n2 but can't quite work it out.
Thanks!
That makes no sense to me at all! I think you meant to ask "why is <0, 0, 1> the normal vector to the x-y plane?" The xy-plane is given by the equation z= 0. Any point in it is of the form (x,y,0) and its "position vector"
is given by <x, y, 0>. In order that <a, b, c> be normal to every such vector we must have the dot product <a, b, c>.<x, y, 0>= ax+ by+ 0= 0. Taking x= 1, y= 0, a(1)+ b(0)+ 0= 0 so we must have a= 0. Taking x= 0, y= 1, a(0)+ b(1)+ 0= 0 so we must have b= 0. A vector is normal to the xy-plane if and only if it is of the form <0, 0, z> and a unit vector normal to the xy-plane is <0, 0, 1>.
"'A'= 2x+ 3y- z= 1" is not a vector at all! If you are asking about the difference between the planes 2x+ 2y- z= 1 and, say, 2x+ 3y- z= 2, they are parallel planes. The first crosses the z axis at (0, 0, -1) (x=y= 0 so the equation becomes 2(0)+ 2(0)- z= 1 or z= -1) and the second at (0, 0, -2).2. what is this significance of vector 'A' = 2x + 3y -z =1 (equaling one)