Answer to Question #102905 in Differential Geometry for vh

Question #102905
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.
1
Expert's answer
2020-02-17T11:25:28-0500

"\\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\n-4y=-2;\\quad\n4z=-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"


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