Find the point, 𝑃∈ℝ3, at which the line 𝑟⃗ (𝑡)=⟨0,−5,−1⟩+𝑡⟨−1,−4,3⟩ intersects the plane 2𝑥+5𝑦+3𝑧=−106.

Answer 1

r(t) = <x(t), y(t), z(t)> = <-t, -5-4t, -1+3t>

So, x = x(t) = -t, y = y(t) = -5-4t, and z = z(t) = -1+3t.

Since 2x + 5y + 3z = -106, we have -2t + 5(-5-4t) +3(-1 + 3t) = -106.

-13t - 28 = -106

-13t = -78 So, t = 6.

The point of intersection is (-t, -5-4t, -1+3t) when t = 6.

So, P = (-6, -29, 17)