Linear Algebra - Find the value of a such that 2 lines lie in a plane

[cris623]

Linear Algebra - Find the value of "a" such that 2 lines lie in a plane

Homework Statement

Consider the lines L1 and L2 with the following equations:
L1: (x y z)=(7 4 −4)+t(2 2 −2)
L2: (x y z)=(a 4 −4)+s(−3 2 −2)
Find the value of a such that L1 and L2 lie in a plane.

They also gave the hint : What must the minimal distance between the two lines be if we are to be sure they lie in the same plane?

Thanks :)

Homework Equations

The Attempt at a Solution

I tried to find the intersection of the two lines
To solve I tried to set up the equations for individual x,y,x coordinates:
L1:
x=7+2t
y=4+2t
z=-4+2t

and L2:
x=a-3s
y=4+2s
z=-4-2s

and then set the equations equal to each other as:

7+2t=a-3s
4+2t=4+2s
-4+2t=4-2s

and then set up a matrix to solve for variables a, s and t

The final time I reduced I got :
(1 -3 -2 | 7)
(0 0 0 | 0)
(0 0 0 | 0)

I'm assuming my thought process was a little off and I'm not too sure where to go from here, and I'm kind of confused how the hint about minimal distance applies to this problem.

Thanks in advance!

 

[Mark44]

Homework Statement

Consider the lines L1 and L2 with the following equations:
L1: (x y z)=(7 4 −4)+t(2 2 −2)
L2: (x y z)=(a 4 −4)+s(−3 2 −2)
Find the value of a such that L1 and L2 lie in a plane.

They also gave the hint : What must the minimal distance between the two lines be if we are to be sure they lie in the same plane?

Thanks :)

Homework Equations

The Attempt at a Solution

I tried to find the intersection of the two lines
To solve I tried to set up the equations for individual x,y,x coordinates:
L1:
x=7+2t
y=4+2t
z=-4+2t

Your equation for z, above, is incorrect. It should be z = -4 - 2t.

and L2:
x=a-3s
y=4+2s
z=-4-2s

and then set the equations equal to each other as:

7+2t=a-3s
4+2t=4+2s
-4+2t=4-2s

The mistake from above is now in your last line, and you have introduced another. The last line should be which should be -4 - 2t = -4 - 2s.

This means that your last two equations are equivalent, so the third is redundant. From the 2nd equation, you can deduce that t = s.

and then set up a matrix to solve for variables a, s and t

The final time I reduced I got :
(1 -3 -2 | 7)
(0 0 0 | 0)
(0 0 0 | 0)

I'm assuming my thought process was a little off and I'm not too sure where to go from here, and I'm kind of confused how the hint about minimal distance applies to this problem.

Thanks in advance!

 

[cris623]

Thanks for the catch on my equation for z. Right before you posted I realized that t=s.

Alright so I switched up my matrix a bit and put "a" on the right hand side and solved for s and t. I got t=(7-a)/-5 and s=(7-a)/-5 which just showed that s and t were equal to each other which I already knew.

Then subbed s and t into the first equation - 7+2t=a-3s hoping I'd be able to solve for a.
Instead I got

-5a+35=14-2a+21-3a which just simplifies to 35=35 and that's pretty obvious.

I'm so lost on how to solve this

 

[Mark44]

Thanks for the catch on my equation for z. Right before you posted I realized that t=s.

Alright so I switched up my matrix a bit and put "a" on the right hand side and solved for s and t. I got t=(7-a)/-5 and s=(7-a)/-5 which just showed that s and t were equal to each other which I already knew.

I solved for a, and got a = 5t + 7, which is equivalent to what you got. t is arbitrary, so pick a value of t to get a value of a.

Then subbed s and t into the first equation - 7+2t=a-3s hoping I'd be able to solve for a.
Instead I got

-5a+35=14-2a+21-3a which just simplifies to 35=35 and that's pretty obvious.

I'm so lost on how to solve this

 

[cris623]

My assignment indicates that there should only be one answer. Even though both you and I came to the conclusion that a varies based on what the value of t is.

 

[cris623]

unless you meant the answer is a=5t+7?

 

[Mark44]

That's what I got for a, but I haven't taken it very much further. I arbitrarily set t = 0 (and s = 0), to get a = 7, but that's the same result as if you merely observed that when a = 7 and s = t = 0, you get the point (7, 4, -4), which is on both lines.

In the hint that's given, the minimal distance they're talking about has to be zero, so we have that, although not in some more elegant manner.

A nicer way to approach the problem is to get an expression that represents the distance between two arbitrary points (one on each line), and then minimize that, either by calculus or by a method that involves vectors and projections and such.