Suppose that the line ℓ is represented by 𝐫(𝑡)=⟨14+2𝑡,17+4𝑡,36+8𝑡⟩ and the plane 𝑃 is represented by 2𝑥+4𝑦+6𝑧=6.

Question

1. Find the intersection of the line ℓ and the plane 𝑃. Write your answer as a point (𝑎,𝑏,𝑐) where 𝑎, 𝑏, and 𝑐 are numbers.

2. Find the cosine of the angle 𝜃

between the line ℓ and the normal vector of the plane 𝑃.

Answer 1

1. To get intersection point we solve system from equation of plane and parametric equation of line: x = 14 + 2t,y = 17 + 4t, z = 36 + 8t

We get 28 + 4t + 68 + 16t + 216 + 48t = 6, 306 = - 68t, t = - 4.5

Now we get point of intersection: (5, - 1, 0).

2.Direction vector of line l is <2, 4, 8> and normal vector to plane is <2, 4, 6>

cosθ =[ (2·2 + 4·4 + 8·6)/(4 + 16 + 64)^1/2·(4 + 16 + 36)^1/2] = 68/4704^1/2 = 0.99146

Answer 2

Hi Kim,

to start conceptually what is happening here is we can imagine a pencil or a needle is poking through a piece of paper and we want to determine the location of the point where they intersect and the angle where the line and the plane meet.

Part 1.) you need to set the x, y, and z components in the line equal to the components in the plane representation and then solve for t.

⟨14+2𝑡,17+4𝑡,36+8𝑡⟩

2𝑥+4𝑦+6𝑧=6

2(14 + 2t) + 4(17 + 4t) + 6(36 + 8t) = 6

28 + 4t + 68 + 16t + 216 + 48t = 6

312 + 68t = 6

t = -4.5

plug the t value into the line equation to get

⟨14+2(-4.5),17+4(-4.5),36+8(-4.5)⟩

This is the point where the plane and the line intersect

⟨5,-1,0⟩

Part 2, the direction vector of the line and the normal vector of the plane are the two components needed to answer the question what is the angle between the line and the plane.

The direction vector of the line can be written in the parametric form like this

⟨14+2𝑡,17+4𝑡,36+8𝑡⟩ = (14,17,36) + (2𝑡,4𝑡,8𝑡) = (14,17,36) + 𝑡(2,4,8)

the vector of the line is (2,4,8)

The normal vector of the plane is (2,4,6)

one way is to use the dot product formula

a•b = |a| |b| cos(ø)

dot product is as follows:

a•b = 2*2 + 4*4 +8*6

a•b = 68

|a| = sqrt (2²+4² +8²)

|a| = 2*sqrt(21)

|b| = sqrt (2²+4² +6²)

|a| = 2*sqrt(14)

68 = 2sqrt(21)*2sqrt(14) * cos(ø)

ø = cos^-1(68 / (2sqrt(21)*2sqrt(14)))

ø ≈ 7.49º