Given two intersecting planes with normals the acute angle between them is found by:
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.
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= n1Ěn2/|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.
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.
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.