tangent planes to level surfaces?

Question

can you help me find the points on the ellipsoid x^2 + 2y^2 + 3z^2 = 1 where the tangent plane is parallel to the plane 3x - y + 3z = 1?

Answer 1

Well, for a point to be on the ellipsoid AND to have partial derivative with respect to x be equal to 3, it has to satisfy two equations in three variables. HOWEVER, just subtract one from the other and you get a quadratic equation in x.

Similar for y and z.

Answer 2

The vector 3i - j + 3k is perpendicular to the plane 3x - y + 3z = 1.

A vector perpendicular to the plane tangent to the ellipsoid at a point (x,y,z) is given by the gradient of F(x,y,z) = x² + 2y² + 3z² - 1, which is 2xi + 4yj + 6zk. For this vector to be parallel to 3i - j + 3k, it must be a multiple of 3i - j + 3k:

2xi + 4yj + 6zk = c(3i - j + 3k) for some number c. So

2x = 3c, 4y = -c, 6z = 3c. Therefore,

c = (2/3)x = -4y = 2z

Use these to replace two of the variables in the equation of the ellipsoid, and solve for the remaining variable.

I get (6/sqrt(41), -1/sqrt(41), 2/sqrt(41)) and
(-6/sqrt(41), 1/sqrt(41), -2/sqrt(41))