Engineering Calculus Notes 47

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online