Question

Consider the curve r(t)=⟨t²−4t,−3t+1,−2t²+5t+4⟩. This is called a planar curve, which means it lies entirely on a single plane. Find an equation for the plane on which this curve lies.

Answer 1

To find an equation for the plane on which the curve lies, we need to find two non-parallel vectors that lie on the plane. We can do this by finding the derivative of the curve and choosing two orthogonal vectors. Finally, we can use the equation of a plane in vector form to find an equation for the plane.

Step-by-step explanation:

To find an equation for the plane on which the curve lies, we need to find two non-parallel vectors that lie on the plane. We can do this by finding the derivative of the curve and choosing two orthogonal vectors. The derivative of the curve r(t) is given by r'(t)=⟨2t-4,-3,10-20t⟩. Choosing two orthogonal vectors, we have v₁ = ⟨2,-3,10⟩ and v₂ = ⟨-4,1,-20⟩.

Now, we can find a normal vector to the plane by taking the cross product of v₁ and v₂. The cross product is n = v₁ x v₂ = ⟨-70,-80,-2⟩. Finally, we can use the equation of a plane in vector form: n·(r-r₀) = 0, where n is the normal vector and r₀ is a known point on the plane, to find an equation for the plane. Substituting the values, we get -70(x-(t²-4t))-80(y-(-3t+1))-2(z-(-2t²+5t+4)) = 0.