Question

Consider the line L(t)=⟨5+t,1+5t⟩. Then:

Answer 1

Answer:

Parallel

Step-by-step explanation:

The answeres are perpinducular becasue of the placement of the lines.

Answer 2

If L(t) = ⟨5 + t, 1 + 5t⟩, then the tangent vector to L(t) is

dL/dt = ⟨1, 5⟩

Any line parallel to L(t) will have the same tangent vector, up to some scalar factor (that is, if the tangent vector is a multiple of ⟨1, 5⟩).

Any line r(t) with tangent vector T(t) = dr/dt that is perpendicular to L(t) will satisfy

T(t) • ⟨1, 5⟩ = 0

• r(t) = ⟨-5, -2t, 1 - 10t⟩ is parallel to L(t) because its tangent vector is

T(t) = ⟨-2, -10⟩ = -2 ⟨1, 5⟩

• r(t) = ⟨1 + 1.5t, 3 + 7.5t⟩ is parallel to L(t) because

T(t) = ⟨1.5, 7.5⟩ = 1.5 ⟨1, 5⟩

• r(t) = ⟨-2 - t, 2 - 2t⟩ is neither parallel nor perpendicular to L(t) because

T(t) = ⟨-1, -2⟩ ≠ k ⟨1, 5⟩

for any real k (in other words, there is no k such that -1 = k and -2 = 5k), and

⟨-1, -2⟩ • ⟨1, 5⟩ = -1 - 10 = -11 ≠ 0

• r(t) = ⟨3 + 15t, -3t⟩ is perpendicular to L(t) because

T(t) = ⟨15, -3⟩

and

⟨15, -3⟩ • ⟨1, 5⟩ = 15 - 15 = 0