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L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

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- Thread starter inknit
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- #1

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L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

- #2

chiro

Science Advisor

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L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

The only way both of these lines can be described by the same variable is if the two lines are linearly dependent.

If this is not the case, you need two variables one for each line.

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- #4

chiro

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I didn't say they don't intersect. If you want to check what happens solve the linear system L1 = L2.

- #5

HallsofIvy

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They don't happen to intersect (and not just "technically") but not because the use the same parameter. A parameter has no meaning outside the equation itself. So as not to confuse yourself, it would be better to change one of them to, say, "s". To determine if they intersect, try to solve x= 11+ 3t= -6+ 4s, y= 7+ t= -2+ 3s, z= 9+ 2t= -7+ 5s for s and t.

L1: x = 11+3t y = 7+t z = 9+2t

L2: x=-6+4t y=-2+3t z=-7+5t

I was given this problem, and technically these lines don't intersect, right?

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