calculus question

Question

For what exact value(s) of k, would the following two planes make an angle of 60°?

plane 1: x+y+kz=6 and plane 2: 2x-y-2z=-5

Answer 1

First off we need to find the point where these 2 lines intersect. Since we are dealing with 3D instead of a regular 2D xy graph, the point of intersection for these 2 planes is actually a line!

So we just need to find the normal vectors of these 2 planes!

Put both of these planes in normal vector form:

1st plane : x+y+(k_)z=6 has normal vector= N_1= i+j+(k_)k (where k_ is the variable we have to find)

2nd plane: 2x-y-2z=-5 has normal vector= N_2= 2(i) - j - 2(z)

And we know the dot product between both normal vectors equals their magnitudes multiplied together also multiplied to cos(60 degrees) so it looks like:

(N_1)(N_2) = (Mag(N_1))(Mag(N_2))(cos(60 degrees)) (where Mag(N_1) is magnitude of N_1 vector and Mag(N_2) is magnitude of N_2 vector)

to solve for variable k_ set cos(60)= [(N_1)(N_2)]/[(Mag(N_1))(Mag(N_2))] so it is:

cos(60)= (1+2(k_))/(3(sqrt(2+(k_)^2))) and solve for k_ !!

so k_ = 0.675 using quadratic formula setting k_ as the x term you have to solve for!!

Answer 2

Normal vector to plane 1: n₁ = < 1, 1, k>

Normal vector to plane 2: n₂ = <2, - 1, - 2>

cos60° = (1·2 - 1·1 - 2k)/[√(2 + k²)·3]; 1/2 = (1 - 2k)/[3√(2 + k²)];

2 - 4k = 3√(2 + k²); 4 - 16k + 16k² = 18 + 9k²; 7k² - 16k - 14 = 0; k = (8 ± √64 + 98)/7 = (8 ± 9√2)/7

This radical equation satisfies only k = (8 - 9√2)/7