Given two intersecting planes with normals the acute angle between them is found by:
.
Problem:
Find an equation of the tangent plane to the ellipsoid x^2+4y^2+z^2=18 at the point (1,2,1) and determine the acute angle that this plane makes with the xy-plane.
Work:
Equation of Tangent Plane is: 2(x-1)+16(y-2)+2(z-1)
Now, this is how I tried to find the acute angle between this plane and the xy-plane:
I used the equation cosθ= n1·n2/|n1||n2|, where n1 and n2 are the normal vector (one of the tangent plane and one of the xy plane).
The normal vector I used for the xy plane was <0,0,1> and the normal vector for the tangent plane is <2,16,2>
Plugging everything into the equation I found that the acute angle is 1.44739 radians or 82.929 degrees.
Can someone verify that this technique is correct? Mainly, I think I need to know if that was the correct normal vector to use for the xy-plane, and if the method I used in general is correct.
Thanks!
1. You can't normalize the dot product of 2 vectors as it's a scalar; that's why I was asking. Are you sure the numerator should have those lines?
2. Can you verify for me that I used the correct normal vector for the xy-plane? It would be much appreciated.
The lines in the numerator are absolute value. The problem asked for the acute angle between the two planes and, for an actute angle, cosine is positive. Taking the absolute value guarentees you get the acute angle and not its complement.
Yes, the z-axis is perpendicular to the xy-plane and so < 0, 0, 1> is a normal vector. Or: the xy-plane is defined by the equation z= 0 or 0x+ 0y+ 1z= 0 so a normal vector is < 0, 0, 1>.Can you verify for me that I used the correct normal vector for the xy-plane? It would be much appreciated.