Final answer:
The function Φ(u,v) parametrizes the plane 2x - y - z = 8 as verified by plugging into the plane equation and satisfies it identically. The tangent vectors Tu and Tv are calculated by taking partial derivatives of Φ, and the normal vector n(u,v) is obtained by their cross product.
Step-by-step explanation:
To show that Φ(u,v)=(7u+4,u−v,13u+v) parametrizes the plane 2x−y−z=8, we can substitute the components of Φ(u,v) into the equation of the plane.
Let x = 7u + 4, y = u - v, and z = 13u + v. Plugging these into the plane equation 2x - y - z = 8 yields:
2(7u + 4) - (u - v) - (13u + v) = 8
14u + 8 - u + v - 13u - v = 8
The v terms cancel out and we get:
14u - u - 13u = 8 - 8
0 = 0
Since the equation is satisfied for all u and v, Φ(u,v) indeed parametrizes the plane.
Next, we calculate the partial derivatives Tu and Tv to find the tangent vectors of the parametrization, and n(u,v) for the normal to the plane.
∂Φ/∂u = Tu = (7, 1, 13)
∂Φ/∂v = Tv = (0, -1, 1)
For the normal vector n(u,v), we take the cross product of Tu and Tv:
n(u,v) = Tu × Tv = (7, 1, 13) × (0, -1, 1) = (14, -7, -7).