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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.)

Respuesta :

Answer:

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⁝š 0.927 = 0.384

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