Answer to Question #102905 in Differential Geometry for vh

Question

Does there exist a plane targent to x^2−2y^2 +2z^2 = 8 and which passes through 2x+3y+2z = 8, x−y+2z = 5? Justify your answer.

Answer 1

$\nabla(F)=(F_x,F_y.F_z)=(2x,-4y,4z)$

Let's find the intersection of two planes, which should be a line with a directional vector equal to the cross product of normal vectors of these planes

$$v=(2,3,2)\times(1,-1,2)=(8,-2,-5)$$

Since the plane we are looking for, passes through that line, it's normal vector of the plane should be parallel to the line's directional vector, therefore

$2x=8;\quad -4y=-2;\quad 4z=-5$

$x=4;\quad y=\frac{1}{2};\quad z=-\frac{-5}{4}$

The equation of the tangent plane

$8(x-4)-2(y-\frac{1}{2})-5(z+\frac{5}{4})=0$