is my intersection of three planes incorrect?

[Alexmcom]

My friend got a different answer than me.

l1=6,-1,0)+t=(3,1,-4) and L2=(4,0,5)+s(-1.1.5)

-1/3, 1/1, 5/4 do not equal to the same answer so it could be skew. We need to check

L1 on the left side

x=6+3t
y=-1+t
z=4t

on the right side L2

x=4-s
y=-s
z=5+5s

For the lines to intersect at a point, there must be one common set of coordinates (x,y,z) for the lines L1 and L2. Therefore, we should be able to find values for S and T that satisfy all three of the following linear equations

6+3t=4-s [1]
-1+t=s [2]
-4t=5+5s [3]

Next step; use the method of elimination to solve this linear system



3t+s=-6+4
3t+s=-2 [4]

-1+t=s
t-s=1 [5]

3t+s=-2 [4]
t-s=1 [5]


multiply step [5] by 3 so we can eliminate t and solve for s

3t+s=-2
3t-3s=3

s=-0.5

plug s into [5]

t-(-.5)=1
t+0.5=1
t=0.5

plug t and s into [3]


-4(0.5)=5+5(-.05)


L1 side is -2

L2 side is 2.5

They do not equal to one another thus the lines are Skew!


He had another answer than me but I feel like I'm wrong. Feel free to correct me!

 

[pka]

My friend got a different answer than me.
l1=6,-1,0)+t(3,1,-4) and L2=(4,0,5)+s(-1,1,5)
x=6+3t
y=-1+t
z=4t

on the right side L2
x=4-s
y=-s
z=5+5s

You have many, many sign errors.

\(\displaystyle \ell_1= \left\{ \begin{array}{l}6+3t\\-1+t\\-4t\end{array} \right.\)-------\(\displaystyle \ell_2 = \left\{ \begin{array}{l}4-s\\s\\5+5s\end{array} \right.\)

Try again.