Find an equation of the plane through the point and perpendicular to the given line?

Find the equation of the tangent plane and the normal line to the surface at the point?

• Find the equation of (a) the tangent plane and (b) the normal line to the given surface at the specified point. 2(x-2)^2 + (y-1)^2 + (z-3)^2 = 10 at (3,3,5)

• Answer:

  Define ƒ(x, y, z) = 2(x-2)² + (y-1)² + (z-3)². Then the gradient of ƒ is: ∇ƒ(x,y,z) = (∂ƒ/∂x, ∂ƒ/∂y, ∂ƒ/∂z) = (4(x-2), 2(y-1), 2(z-3)). The vector ∇ƒ(3,3,5) is normal to the surface ƒ(x,y,z) = 10 at (x,y,z) = (3,3,5). Compute it: ∇ƒ(3,3,5) = (4(3-2), 2(3-1), 2(5-3)) = (4, 4, 4). (a). The tangent plane is normal to ∇ƒ(3,3,5), so for (x,y,z) in the plane, ∇ƒ(3,3,5)•(x-3,y-3,z-5) = 0. (4,4,4)•(x-3,y-3,z-5) = 0. 4(x-3)+4(y-3)+4(z-5) = 0. (b). The normal line is parallel to the vector ∇ƒ(3,3,5), so for any (x,y,z) in the line, there is a real number t with (x-3,y-3,z-5) = t∇ƒ(3,3,5) = t(4,4,4). This is a parametric equation for the line (with parameter t). If you prefer to express the line as the intersection of two planes, you can eliminate t, and express y and z in terms of x: 4t = x - 3. t = 1/4(x - 3). So y - 3 = 4t = 4/4(x - 3), so y = x, and z - 5 = 4t = 4/4(x - 3) = x - 3, so z = x + 2.