Find the point at which the line ⟨1,0,−4⟩+𝑡⟨1,3,−5⟩ ⟨ 1 , 0 , − 4 ⟩ + t ⟨ 1 , 3 , − 5 ⟩ intersects the plane 4𝑥−3𝑦−2𝑧=57 4 x − 3 y − 2 z = 57 .

Question

Find the point at which the line  intersects the plane .

Answer 1

x = 1 + t; y = 3t; z = - 4 - 5t.

Substitute in plane equation: 4(1 + t) - 3(3t) - 2(- 4 - 5t) = 57; 4t - 9t + 10t = 57 - 12; 5t = 45; t = 9.

So, x = 10, y = 27, z = - 49.

Intersection point: (10, 27, - 49)