Final answer:
The equation of the normal plane can be found using the components of the tangent vector as the normal vector and a point from the position vector. The general form is (x-x0)a + (y-y0)b + (z-z0)c = 0, where (a, b, c) is the normal vector. Without the specific value of t, we cannot match the equation to the provided answer choices.
Step-by-step explanation:
The student wants to find the equation of the normal plane given a tangent vector and a position vector r. The tangent vector provided is (1, t, 0.5t²), which implies a directional movement along the curve. The normal plane to a curve at a certain point is a plane that is perpendicular to the tangent vector at that point.
To find the normal plane, we can use the fact that the normal vector to the plane is given by the tangent vector to the curve. Therefore, the normal vector can be written directly from the components of the tangent vector, which are (1, t, 0.5t²). The equation of a plane is given by (x-x0, y-y0, z-z0) ⋅ (a, b, c) = 0, where (x0, y0, z0) is a point on the plane (which can be taken from the position vector r), and (a, b, c) is the normal vector to the plane.
Using the position vector r=(t, 0.5t², t³/6) at a point (say when t=1), we get the point (1, 0.5, 1/6). Using this point and the components of the tangent vector as the normal vector, we get the equation of the normal plane by the dot product:
(x-1)⋅ 1 + (y-0.5)⋅ t + (z-1/6)⋅ 0.5t² = 0
Which simplifies to:
(x-1) + t(y-0.5) + 0.5t²(z-1/6) = 0
To find the specific equation corresponding to the given answer choices, we would need to know the specific value of t at the point we're interested in. Without this information, we cannot correctly match to one of the given options a), b), c), or d).