Final answer:
To find the osculating plane's equation using a binormal vector and a position vector requires additional information about the curve or a specific value of the parameter t, which is not provided in the question. Therefore, a precise equation for the osculating plane cannot be determined without such information.
Step-by-step explanation:
The question pertains to finding the osculating plane of a curve given a position vector r and a binormal vector Binormal. The osculating plane is defined at a certain point on a curve and is determined by the direction of the curve's tangent vector T, normal vector N, and binormal vector B. In vector calculus, given a binormal vector and a point on the curve (through the position vector), the osculating plane's equation can be found as the plane that is perpendicular to the binormal vector at that point.
To find the equation of the osculating plane, we must compute the cross product between the curve's tangent and normal vector, which results in the direction of the binormal vector. Given the binormal vector (0.5t², -t, 1), and assuming that this vector is perpendicular to the osculating plane, we can set up the plane equation in the form ax + by + cz = d, where a, b, and c are the components of the binormal vector and d is a constant which can be found by substituting the point from vector r.
However, without more information about the curve itself or the particular value of t at which the osculating plane is desired, we cannot provide a definitive answer for d in this case. It is worth noting that the answer options given don't seem to correlate with the normal vector provided; they may correspond to a different set of information about the tangent or normal vectors.