Find Line Intersecting Line 1 & 2 at 90°: Vector, Parametric & Symmetric Eqs

[Failure007]

Homework Statement

Find the vector, parametric and symmetric equations of a line that intersect both line 1 and line 2 at 90°.
L1 :

x = 4 + 2t
y = 8 + 3t
z = -1 - 4t
L2 :

x = 7 - 6t
y = 2+ t
z = -1 + 2t

Homework Equations

$$\vec{a}$$₁ - $$\vec{a}$$₂ + $$\vec{b}$$₁t - $$\vec{b}$$₂s

The Attempt at a Solution

=[4,8,-1] - [7,2,-1] + [2t, 3t, -4t] - [-6s, 1s, 2s]

=[-3, 6, 0] + [(2t + 6s), (3t-1s), (-4t-2s)]

=[(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)]

=[2, 3, -4] . [(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)]

=(-6 + 4t + 12s) + (18 + 9t -3s) + (0 +16t + 8s)

=12 + 29t +17s -------(equation 1)

=[-6, 1, 2] . [(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)]

=(18 -12t -36s) + (6 + 3t - s) + (0 -8t -4s)

=24 -17t -41s ----------(equation 2)

29t + 17s = -12
-17 -41s = -24

-493t -1189s = -696
-493t -289s =204
-900s = -900

Therefore, s = 1

29 + 17(1) = -12
29t = -29

Therefore, t=-1

At these values, the dot product of

$$\vec{b}$$₁ . ($$\vec{a}$$₁ - $$\vec{a}$$₂ + $$\vec{b}$$₁t - $$\vec{b}$$₂s)

and

$$\vec{b}$$₂ . ($$\vec{a}$$₁ - $$\vec{a}$$₂ + $$\vec{b}$$₁t - $$\vec{b}$$₂s)

both equal 0, meaning this should be perpendicular to both lines.
---------------------------------------------------------------------------------------------------
I'm honestly not even sure if I'm on the right track with this but it seemed like it was working up until this point, however I'm not really sure how to derive the vector equation from this given information, so if anyone at all has some info that would point me in the right direction it would be greatly appreciated. Thanks to anyone and everyone willing to help.

 

[dynamicsolo]

You may have confused yourself by writing equals signs on every single line; you should only use that symbol to connect statements that actually are equal. You may want to describe things this way:

You have the lines

L1 :
x = 4 + 2t
y = 8 + 3t
z = -1 - 4t

L2 :
x = 7 - 6s
y = 2+ s
z = -1 + 2s

A line connecting a point on L1 with a point on L2 satisfies

[4,8,-1] - [7,2,-1] + [2t, 3t, -4t] - [-6s, 1s, 2s]

= [-3, 6, 0] + [(2t + 6s), (3t-1s), (-4t-2s)]

= [(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)] .

A line through these points and perpendicular to both L1 and L2 must simultaneously satisfy

[2, 3, -4] . [(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)]

= (-6 + 4t + 12s) + (18 + 9t -3s) + (0 +16t + 8s)

=12 + 29t +17s = 0 -------(equation 1)

AND

[-6, 1, 2] . [(-3 + 2t +6s), (6 + 3t - s), (0 - 4t -2s)]

=(18 -12t -36s) + (6 + 3t - s) + (0 -8t -4s)

=24 -17t -41s = 0 ----------(equation 2)

The values of t and s which satisfy both of these equations are given by

29t + 17s = -12
-17t -41s = -24

-493t -1189s = -696
-493t -289s =204
-900s = -900

[Ah, here's a faster way to do this: the second equals equals something exactly twice as big as the first, so

2 · (29t + 17s) = -17t - 41s , or

58t + 34s = -17t - 41s , so 75t = -75s , or t = -s .

Putting this result into either equation shows that
t = -s = -1 . ]

Now then, you have the values of the parameter for a point on each line. Putting the appropriate parameter into the equations for L1 or L2 gives you the coordinates of the point on each line, for which the segment linking them is perpendicular to both L1 and L2. Finally, find the equation(s) of the line containing both of those points.

Your ideas for solving the problem are basically sound. And when you find the vector for the solution line, you will indeed find that its dot products with the vectors for L1 and L2 are both zero.

 

[Failure007]

First of all thanks for responding, what I'm still sort of confused about is if I plugged those parameters into their respective equations i.e.

[x,y,z] = [4,8,-1] + -1[2,3,-4] = [2, 5, 3] for line 1

[x,y,z] = [7,2,-1] + 1[-6,1,2] = [1, 3, 1] for line 2

would that mean that those aren't the correct parameters or do I have more work to do? I'm not really sure where I would begin.

 

[dynamicsolo]

What you now have are the coordinates of a point on each line,

(2, 5, 3) on line 1 and (1, 3, 1) on line 2 .

That is where the parameters t and s are telling us those points are. You set up the system of two equations to solve for the point on L1 and on L2 such that the line passing through both intersects both L1 and L2 at right angles.

You now want to find the equation for the line which contains both (2, 5, 3) and (1, 3, 1) -- and you're done! Check the dot products of that line's vector with the vectors for L1 and L2 and see what you get...

 

[Failure007]

Makes perfect sense, thanks so much.