Unformatted text preview: ℓ 2 −→ p 2 (0) = (1 , , 1) lies on ℓ 1 . This means we need to see if for some value of t we have −→ p 2 (0) = −→ p 1 ( t ), or 1 = 1 − 3 t 0 = − 2 − 2 t 1 = 3 + t. It is easy to see that the ²rst equation requires t = 0, the second requires t = − 1, and the third requires t = − 2; there is no way we can solve all three simultaneously. It follows that ℓ 1 and ℓ 2 are distinct, but parallel, lines. Now, consider a third line, ℓ 3 , given by x = 1 +3 t y = 2 + t z = − 3 + t....
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 Fall '08
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 Calculus, ﬁrst equation, coordinate form

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