x + y + z = 2, x + 3y + 3z = 2 (a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.) (x (t), y (t), z (t)) = (b) Find the angle between the planes. (Round your answer to one decimal place.)

a) L (t) = (2,0,0) + (0,-1,1) * t

b) θ = 0.384 rad

Step-by-step explanation:

a) for the planes

x + y + z = 2

x + 3y + 3z = 2

then substracting the first equation to the second

2y+2z=0 → y = - z

replacing in the first equation

x + (-z) + z=2 → x=2

thus choosing t=z as parameter

x=2

y=-t

z=t

or

L (t) = (2,0,0) + (0,-1,1) * t

b) the angle can be found through the dot product of the normal vectors to the plane

n1*n2 = (1,1,1) * (1,3,3) = 1*1+1*3+1*3 = 7

|n1| = √ (1²+1²+1²) = √3

|n2| = √ (1²+3²+3²) = √19

since

n1*n2 = |n1|*|n2|*cos θ

cos θ = n1*n2 / |n1|*|n2| = 7 / (√3*√19) = 0.927

thus

θ = cos⁻¹ 0.927 = 0.384