Roselyn Daniel
2022-08-01
Answered

Determine if there is a plane in ${\mathbb{R}}^{3}$ which contains both lines ${l}_{1}\text{}and\text{}{l}_{2}$ 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.

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Julianna Bell

Answered 2022-08-02
Author has **19** answers

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.

$<2,1,3>\times <3,3,4>=\left|\begin{array}{ccc}i& j& k\\ 2& 1& 3\\ 3& 3& 4\end{array}\right|=<-5,1,3>$

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

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.

$<2,1,3>\times <3,3,4>=\left|\begin{array}{ccc}i& j& k\\ 2& 1& 3\\ 3& 3& 4\end{array}\right|=<-5,1,3>$

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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