Final answer:
The function Φ(u,v)=(7u+4,u−v,13u+v) indeed parametrizes the plane 2x-y-z=8. The tangent vectors are Tu = (7, 1, 13) and Tv = (0, -1, 1), and the normal vector to the plane is n(u, v) = (1, -7, -7).
Step-by-step explanation:
To show that Φ(u,v)=(7u+4,u−v,13u+v) parametrizes the plane 2x-y-z=8, we substitute the corresponding coordinates from Φ into the equation of the plane.
- Let x = 7u + 4
- Let y = u - v
- Let z = 13u + v
Substitute into the plane's equation: 2(7u + 4) - (u - v) - (13u + v) = 8. This simplifies to 14u + 8 - u + v - 13u - v = 8, which further simplifies to 8 = 8. Thus, Φ parametrizes the plane.
To find Tu and Tv, we compute the partial derivatives of Φ with respect to u and v respectively.
Tu = Φu = ∂Φ/∂u = (7, 1, 13)
Tv = Φv = ∂Φ/∂v = (0, -1, 1)
To find the normal vector n(u, v) to the plane, we take the cross product of Tu and Tv.
n(u, v) = Tu × Tv = (7, 1, 13) × (0, -1, 1) = (1, -7, -7)