Engineering Calculus Notes 47

Engineering Calculus Notes 47 - ℓ 2 −→ p 2(0 =(1 1...

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1.3. LINES IN SPACE 35 or, in coordinate form, x = 1 3 t y = 2 2 t z = 3 + t while the second, 2 , is given in coordinate form as x = 1 +6 t y = 4 t z = 1 2 t. We can easily read oF from this that 2 has parametrization −→ p 2 ( t ) = (1 , 0 , 1) + t (6 , 4 , 2) . Comparing the two direction vectors −→ v 1 = 3 −→ ı 2 −→ + −→ k −→ v 2 = 6 −→ ı + 4 −→ 2 −→ k we see that −→ v 2 = 2 −→ v 1 so the two lines have the same direction—either they are parallel, or they coincide. To decide which is the case, it su±ces to decide whether the basepoint of one of the lines lies on the other line. Let us see whether the basepoint of
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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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