Question

Where does the normal line through to the paraboloid z=3x²+y² at the point (1,2,7) intersect the paraboloid a second time? Let's do this in steps The normal vector to the paraboloid at (1,2,7) is given by ⟨,−1⟩. Using this normal vector the most natural equation for the normal line is: L(t)= This normal line intersects the paraboloid at: t=0 and t= And therefore the line intersects the paraboloid again at:

Answer 1

To determine the second intersection point of the normal line with the paraboloid, find the gradient at the given point, use it to form the normal line equation, substitute into the paraboloid's equation, solve for t, then substitute back to find the coordinates.

Step-by-step explanation:

The question is about finding the point where the normal line to a paraboloid at a given point intersects the paraboloid again. To start, we need to find the gradient of the paraboloid. For the paraboloid given by z = 3x² + y², the gradient at point (x, y, z) is ∇z = ⟂(6x, 2y, -1). The negative sign in the z-component recognizes that the normal to the surface points in the opposite direction of the gradient.

At the given point (1, 2, 7), we have ∇z = ⟂(6, 4, -1). This gradient vector is used to form the equation of the normal line L(t) = (1, 2, 7) + t(6, 4, -1). To find where this line intersects the paraboloid a second time, we substitute the expressions for x, y, and z from the normal line into the equation of the paraboloid and solve for t.

The resulting equation is 3(1 + 6t)² + (2 + 4t)² = 7 - t. This quadratic equation in t will have two solutions, one of which is t = 0, representing the initial point of intersection. Solving for the second value of t will give us the second intersection point. Finally, we substitute this value back into the equation of the normal line to find the coordinates of the second intersection.