Final answer:
To find the equation for the tangent plane at the point P0(1,2,1) on the surface x^2+3y^2+4z^2=17, we first find the gradient of the surface ∇f(P0) = (2, 12, 8), then write the equation of the tangent plane as 2x + 12y + 8z - 26 = 0. The equation for the normal line can be written as x = 1 + 2t, y = 2 + 12t, z = 1 + 8t where t is a parameter representing the distance from the point P0.
Step-by-step explanation:
To find the equation for the tangent plane at the point P0(1,2,1) on the surface x^2+3y^2+4z^2=17, we first need to find the gradient of the surface. The gradient is a vector that points in the direction of the steepest increase in the function. In this case, the gradient is given by ∇f = (2x, 6y, 8z). Evaluating the gradient at the point P0 gives us ∇f(P0) = (2, 12, 8).
The equation for the tangent plane can then be written as:
(x - 1)(2) + (y - 2)(12) + (z - 1)(8) = 0
Simplifying the equation gives us the equation of the tangent plane: 2x + 12y + 8z - 26 = 0.
To find the equation for the normal line at the point P0, we use the direction vector of the line, which is the gradient ∇f(P0) = (2, 12, 8). The equation of the line can be written as:
x = 1 + 2t, y = 2 + 12t, z = 1 + 8t
where t is a parameter that represents the distance from the point P0 along the line.