Equations of Lines

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K_Swiss

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WHERE DID I GO WRONG?


Question:

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]

----------------

My Attempt:

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 (1) 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, POI = (-1/3 , 3/2 , 0)


----------------


Textbook Answer:
(-1,1,1)
 
K_Swiss said:
WHERE DID I GO WRONG?


Question:

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]

----------------

My Attempt:

d_1 = (4,1,1)
d_2 = (-3,2,-1)

Therefore, d_1 is not parallel to d_2

>> RIGHT.

(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
================= LOOKS GOOD UP TO HERE. =============
Now you substitute your s = 1, t = -1 into your equations for x,y,z.
Of course you have two equations for each, such as:

x = 3 + 4t
y = 2 + t
z = 2 + t

x = 2 - 3s
y = - 1 + 2s
z = 2 - s

which, I think, you got already. But BOTH equations for x will give you the same answer:

x = 3 + 4(-1) = -1
x = 2 - 3(1) ALSO = -1

Likewise your two equations for y and for z will give you the same answers--

y = 1, z = 1.



================ I DON'T KNOW WHAT THIS PART DOES. ==========
Verify (1) using "t" and "s":
LS:
= 4(-1) + 3(1)
= -4 + 3
= -1
RS:
= -1


Therefore, LS = RS

POI:

x:
3 = 2 - 3s << This looks wrong.
-1/3 = s

y:
2 = -1 + 2s
3/2 = s

z:
2 = 2 - s
0 = s

Therefore, POI = (-1/3 , 3/2 , 0)


----------------


Textbook Answer:
(-1,1,1)
Click to expand...
 
t = -s, t = -1, s = 1, hence POI = (-1,1,1)

Check: x = 3+4t, x = -1, x = 2-3s, x = -1

y = 2+t, y = 1, y = -1 +2s, y = 1

z = 2+t, z =1, z = 2-s, z = 1
 
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