Final answer:
The Cartesian equation of the plane normal to the vector rt at t=0 and containing the point (1,1,1) is 2x + 6z = 8.
Step-by-step explanation:
To find the Cartesian equation of the plane containing the point (1, 1, 1) and normal to the vector rt=2+t,3t,6, we need to understand a basic property of planes in a three-dimensional Cartesian coordinate system. A plane equation can be written in the form Ax + By + Cz = D, where (A, B, C) are the components of a normal vector to the plane, and D is a constant that can be determined using a known point on the plane.
In this case, the given normal vector is rt(0)=(2,0,6), as we evaluate the vector rt at t=0. The normal vector determines the coefficients A, B, and C in the plane's equation. Therefore, the equation of the plane can be expressed as 2x + 0y + 6z = D. To find D, we substitute the point (1, 1, 1) into the equation, yielding 2(1) + 0(1) + 6(1) = D, which simplifies to D=8.
The Cartesian equation of the plane is therefore 2x + 6z = 8.