Final answer:
To find the point where the curve intersects the xz-plane, set the y-coordinate to 0 and solve for t. Plug the value of t back into the equations to find the coordinates. To find the equation of the normal plane to c at a given point, find the derivative of the curve equations and evaluate it at the given point. Take the cross product of the derivatives to find the normal vector, and use it to write the equation of the normal plane.
Step-by-step explanation:
To find the point where the curve intersects the xz-plane, we need to set the y-coordinate to 0. From the given equation y = 2t - 1, we can solve for t as follows: 0 = 2t - 1, t = 0.5. Plugging this value of t back into the equations, we get x = 2(0.5) - 3 = 0, and z = ln(0.5) ≈ -0.6931. Therefore, the point where c intersects the xz-plane is (0, 0, -0.6931).
To find the equation of the normal plane to c at the point (1, 1, 0), we need to find the derivative of the curve equations and evaluate it at t = 1. Differentiating the given equations with respect to t, we get dx/dt = 2 and dy/dt = 2. Taking the cross product of the derivatives, we get the normal vector as (0, 0, -4). Since the point (1, 1, 0) lies on the curve, the equation of the normal plane is ax + by + cz = 0, where a = 0, b = 0, and c = -4. Therefore, the equation of the normal plane is -4z = 0.