# Determine if there is a plane in mathbb(R)^3 which contains both lines l_1 and l_2 where l_1 : x= 2t + 3, y= t +3, z = 3t +8 and l_2: x=3s - 2, y = 3s - 1, and z = 4s+1.

Determine if there is a plane in ${\mathbb{R}}^{3}$ which contains both lines where: ${l}_{1}:x=2t+3,y=t+3,z=3t+8$ and ${l}_{2}:x=3s-2,y=3s-1$, and z = 4s+1.
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Julianna Bell
In order for a plane to contain both lines then the lines musteither be parallel or intersect...
the direction of L1 is <2, 1, 3> and the direction of L2 is <3, 3, 4>
since there is no k value such that...
$<2,1,3>=k\ast <3,3,4>\frac{2}{3}\ne \frac{1}{3}\ne \frac{3}{4}$
then the lines are not parallel.
determine if they intersect by equating like components (x =x, y = y, z = z).
x: 2t+3=3s-2...(1)
y: t+3=3s-1...(2)
z: 3t+8=4s+1...(3)
solving (2) for t yields t = 3s - 4
sub into (1)... 2(3s - 4) + 3 = 3s - 2
s = 1
sub into (2).. t + 3 = 3(1) - 1
t = -1
test s = 1 and t = -1 in (3)... 3(-1) + 8 = 4(1) +1
therefore the lines do intersect and thus also lie in a common plane.
the normal direction of that plane is orthogonal to both lines so use cross product to find it.
$<2,1,3>×<3,3,4>=|\begin{array}{ccc}i& j& k\\ 2& 1& 3\\ 3& 3& 4\end{array}|=<-5,1,3>$
we just need a point from either line to plug into theequation for a plane now...choosing t = 0 from L1 yields the point(3, 3, 8) so the equation of the plane is.
-5(x-3)+1(y-3)+3(z-8)=0
-5x+y+3z-12=0
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