Question

How would you parameterize the plane through the point (4,−2,5) with the normal vector ⟨−4,2,−5⟩?

a) r = (4−4t, −2+2t, 5−5t)
b) r = (4−t, −2+t, 5−t)
c) r = (4t, −2t, 5t)
d) r = (4+t, −2−t, 5+t)

Answer 1

The correct parameterization of the plane through point (4, −2, 5) with normal vector ⟨−4, 2, −5⟩ requires two independent direction vectors on the plane, but none of the provided options represent a plane parameterization since they suggest a line with only one parameter. More information is needed to provide a correct answer.

Step-by-step explanation:

To parameterize the plane through the point (4, −2, 5) with the normal vector ⟨−4, 2, −5⟩, we need to find a vector equation that represents all points (x, y, z) on the plane.

A plane can be parameterized by a point on the plane (called the reference point) and two independent direction vectors that lie on the plane. Since we are given a normal vector, we need direction vectors that are perpendicular to the normal vector.

The generic form of the parameterization of a plane is r(t, s) = r0 + t*v1 + s*v2, where r0 is the reference point, t and s are parameters, and v1 and v2 are direction vectors on the plane.

However, the given options for parameterization suggest a line rather than a plane, since they do not include two parameters.

As none of the options correctly parameterizes a plane, the correct parameterization cannot be determined without further information to identify appropriate direction vectors for the plane.