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Perpendicular, Parallel and Skew Lines in Space - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
How can you tell if two lines intersect in space? Let’s take a look at an example. I have two lines here L1 and L2. L1 is x,y,z equals 0, 3, -4 plus t times the direction vector 2, 1, 3 . And L2 is x,y,z equals 5, 1, 2 plus s times the direction vector 1, 2, 4. There are different lines so use different parameters t and s.
To find out where they intersect, I’m first going write their parametric equations. So starting with L1. L1 is going to be x equals 0 plus 2t, x equals 2t. Y equals 3 plus t, and z equals -4 plus 3t.
What about L2? X equals 5 plus s, y equal 1 plus 2s, and z equals 2 plus 4s. So to find out where they intersect, I’m actually going to set the x equations equal, the y equations equal, and the z equations equal. Setting the x equations equal, I get 2t equals 5 plus s. Setting the y's equal, I get 3 plus t equals 1 plus 2s. And the z’s, I get -4 plus 3t equals 2 plus 4s. Let me simplify these a little bit. I’ll get the variables on one side and the constants on the other. And I get minus s plus 2t equals 5. And here I get minus 2s plus t, and I have to subtract 3 from both sides so I get equals -2. And here I get minus 4s plus 3t, and I have to add 4 to both sides. So I get equals 8, I’m going to call this equation 1, this one 2, and this one 3.
Notice if I multiply the top equation by -2 and add it to the second, I’ll eliminate the s’s. So -2 times 1 plus equation 2. This just tells you what I’m doing; multiplying equation 1 by -2 and adding it to 2. I get positive 2s plus -2s is 0, and I get -4t plus t is -3t equals. And then minus 10, minus 2, -12.
So from this equation, I get t equals 4. Let me also multiple this equation by -2, and add it to this one. So I’m multiplying equation 2 by -2 and adding it to equation 3, and I’ll get +4s minus 4s is 0. I get minus 2t, plus 3t is t, so plus t equals, and I’ll get 4 plus 6, 10. I get t equals 10. Now t can't both be 4 and 10. So right here we have a contradiction and that means that these two lines can never intersect. Lines do not intersect.
Because they don’t intersect, I want to find out if they're parallel. So the question, are they parallel? Remember how to test for parallel lines. The direction vectors have to be scalar multiples of one another. So let’s see if this direction vector is a scalar multiple of this one. So can you find a k value so that 1, 2, 4 is a scalar multiple of 2, 1, 3? Well we could try k equals 1/2, and if that doesn’t work then they are not parallel.
So let me to just evaluate 1/2 times 2, 1, 3. And I get 1/2 and 3/2. And that is not the same as 1, 2, 4. So these lines are not parallel and that means they are skew.
Remember skew lines are two lines in space, that never meet but aren’t parallel. They just go right by each other like this.
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