haydenamrhein1693
10.02.2020 •
Mathematics
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.)
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Ответ:
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
Ответ:
m = 2
Step-by-step explanation: Coz I juss know