[b]WHERE DID I GO WRONG?[/b]
[b]Question:[/b]
Find the intersection of each pair of lines. If they do not meet, determine whether they are parallel and distinct or skew.
n)
[(x - 3)/4] = [y - 2] = [z - 2]
[(x - 2)/-3] = [(y + 1)/2] = [(z - 2)/-1]
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[b]My Attempt:[/b]
d_1 = (4,1,1)
d_2 = (-3,2,-1)
Therefore, d_1 is not parallel to d_2
(1) 3 + 4t = 2 - 3s -> 4t + 3s = -1
(2) 2 + t = -1 + 2s -> t - 2s = -3
(3) 2 + t = 2 - s -> t + s = 0
Find "s", using substitution, sub (3) into (2).
s = 1
Find "t" using substitution, sub "s" into (3).
t = -1
Verify (3) using "t" and "s":
LS:
= 4(-1) + 3(1)
= -4 + 3
= -1
RS:
= -1
Therefore, LS = RS
POI:
x:
3 = 2 - 3s
-1/3 = s
y:
2 = -1 + 2s
3/2 = s
z:
2 = 2 - s
0 = s
Therefore, [b]POI = (-1/3 , 3/2 , 0)[/b]
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[b]Textbook Answer:[/b]
[b](-1,1,1)[/b]
2 answers
You used the 2nd and 3rd equation to find s and t
You have to verify in the equation that was not used, and that would be the first
in the first
LS = 4t+3s
= 4-3
= -1
RS = =1 which is NOT the left side
So they do not meet, and since they are not parallel they must be skew lines