Question

[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]

Answer 1

The problem lies in your verification

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

Answer 2

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