Find the plane determined by the intersecting lines (Calculus III Help)?
Hello, I need some help with this problem. My calculations are not working out. Please show your steps! Thank you so much for your assistance.
I need to find the plane determined by the intersection of the following lines:
L1:
x = -1 + t
y = 2 + t
z = 1 - t
L2:
x = 1 - 4s
y = 1 + 2s
z = 2 - 2s
The answer I have is: y + z = 3.
Thanks again for your help!
2 Answers
x=-1+t
y=2+t
z=1-t
——–
x=1-4s
y=1+2s
z=2-2s
———-
-1+t=1-4s =>t=-4s+2
2+t=1+2s
4-4s=1+2s
6s=3 =>s=1/2 =>t=0
verify:
1-t=2-2s
1-0=2-1
1=1 true
the point of intersection is (-1,2,1)
———–
the parametric form of the lines are:
L1=(-1,2,1)+t(1,1,-1)
L2=(1,1,2)+s(-4,2,-2)
take the cross product of:
<1,1,-1>*<-4,2,-2> = <0,6,6>
as it passes through (-1,2,1) then:
0(x+1)+6(y-2)+6(z-1)=0
6y+6z=18
y+z=3
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