Equation of a tangent Plane

[ryan8888]

Given: let f (x,y,z) = x² + y² + z² = 2y - 3x and g(x,y,z) = 3x + y² - z²

A. Find an equation for the tangent plane for the surface g(x,y,z) = 9 at the point (3, -1, 1)
B. Find the line tangent to the intersection of the surfaces f(x,y,z) = 0 and g(x,y,z) =9 at the point (3,-1,1)

A. I feel like I am reading into this part too much. In part A do I need to simply utilize g(x,y,z) and take the gradient of the g(x,y,z) and then subsitute the point (3,-1,1) into the gradient function and then finally use the information found from this in the form a(x-x₀) + b(y-y₀) + c(z-z₀) = 0 to get the equation of the tangent plane.

Or do I need to utilize both f(x,y,z) and g(x,y,z). O would take the gradient of both functions and compute the cross product to get the normal vector, then substitute the point (3,-1,1) into the normal vector and input that information into the form: a(x-x₀) + b(y-y₀) + c(z-z₀) = 0 to solve for the equation of the tangent plane.

Thanks for any help!

 

[HallsofIvy]

a) Yes if a surface is given by g(x,y,z)= constant, then $\nabla f= grad f$ is perpendicular to the surface and so a normal to the tangent plane.
Of course, a plane containing point $(x_0, y_0, z_0)$ having normal vector <A, B, C> has equation $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. This problem has nothing to do with the function f.

b) A line tangent to the curve of intersection of two surfaces must lie in the tangent planes of both surfaces- and therefore is perpendicular to the normal to both surfaces at that point. Find the cross product of $\nabla f$ and $\nabla g$, a vector in the direction of that line. Of course, the line through point $(x_0, y_0, z_0)$ in the direction of the vector <A, B, C> is given by the parametric equations $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$.

 

[ryan8888]

Awsome. Thanks for the help. I figured that the first portion of the question the f(x,y,z) had nothing do do with determing the equation of the plane. I came up with 3x - 2y - 2z = 9 for the equation of the plane that is tangent to g(x,y,z).

I do have another question for part 2. I have taken the gradients of both f(x,y,z) and g(x,y,z) and come up with grad f = (2x-3, 2y + 2, 2z) and grad y = (3, 2y, -2z) should I substitute my point (3, -1, 1) into the gradients and then take the cross product or should I do the cross product and then sub the values for my point in?

a) Yes if a surface is given by g(x,y,z)= constant, then $\nabla f= grad f$ is perpendicular to the surface and so a normal to the tangent plane.
Of course, a plane containing point $(x_0, y_0, z_0)$ having normal vector <A, B, C> has equation $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. This problem has nothing to do with the function f.

b) A line tangent to the curve of intersection of two surfaces must lie in the tangent planes of both surfaces- and therefore is perpendicular to the normal to both surfaces at that point. Find the cross product of $\nabla f$ and $\nabla g$, a vector in the direction of that line. Of course, the line through point $(x_0, y_0, z_0)$ in the direction of the vector <A, B, C> is given by the parametric equations $x= At+ x_0$, $y= Bt+ y_0$, $z= Ct+ z_0$.