Check my working for finding if vectors are parallel ?

  • Thread starter brandy
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  • #1
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Homework Statement


there is a line through A : (1;-2; 1) and B : (0; 2; 3)
and the line through C : (4; 1;-2) and D : (2; 2; 2)
are they parallel?

The Attempt at a Solution


i said no they weren't.
i had the lines as being:
(-t+1)i+(4t+2)j+(2t+1)k
and (-2t+4)i+(t+1)j+(4t-2)k

and then i took the cross product assuming constants don't count.
and got a determinant-y thing of 21 which is not 0 - why i said it wasn't parallel.
is my working correct?
 

Answers and Replies

  • #2
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Let's simplify because this is straight forward without the parametrization.

Define a vector between A and B (taking the components and subtracting them) and the same for C and D. Then take the coefficients of one vector and see if we can multiply all of them by one number to get the other.

If you can't find this number, then we're out of luck.
 
  • #3
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the question asks for the equation of the lines. so i did that.
i posted to see if i made any errors
are my answers correct or are they not?
 
  • #4
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Ok, we can use the same idea then. See if you can find a number that gets from one line to the other. I wouldn't use a cross product, but if I did, then I'd need to keep the constants.

I'll also hazard that posting sort of opens you up to questions about process and understanding. The important thing here is that we get the process right. You'll probably never see this exact problem again, but you might see a variation. So, seeing if you just "made any errors" is a bit of a misnomer. Life is not about the right answer.

So, no. Its not correct if you left out the constants in the cross product. You can redo that with a half page at best of math or you can use the simpler version from two posts ago.
 
  • #5
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hmm.
well, we're supposed to use the stuff we are learning about right now.

i was told by my teacher that i could use either dot or cross on the lines.
he did an example for an equation for a line with two variabkes (s and t) and found the normals (another equation with i,j and k in it like my previous ones)
i figured that because there was no 2 variables and it was kind of the same form i didnt need to find the normals. plus i remember him saying something about the constants and how they didnt affect it.
for his example he used dot product, i wanted to try cross product. so. this is my thought proccess.
 
  • #6
tiny-tim
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hi brandy! :smile:
there is a line through A : (1;-2; 1) and B : (0; 2; 3)

(-t+1)i+(4t+2)j+(2t+1)k

(4t minus 2) :wink:
 
  • #7
HallsofIvy
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Homework Statement


there is a line through A : (1;-2; 1) and B : (0; 2; 3)
and the line through C : (4; 1;-2) and D : (2; 2; 2)
are they parallel?

The Attempt at a Solution


i said no they weren't.
i had the lines as being:
(-t+1)i+(4t+2)j+(2t+1)k
and (-2t+4)i+(t+1)j+(4t-2)k
So they have "direction vectors" -i+ 4j+ 2k and -2i+ j+ 4k. Is one of those a multiple of the other?

and then i took the cross product assuming constants don't count.
and got a determinant-y thing of 21 which is not 0 - why i said it wasn't parallel.
is my working correct?
Took the cross product of what? I only ask in order to encourage you to think and write more clearly. Yes, the cross product of the two direction vectors would have to be 0 for the lines to be parallel. But I think it is simpler to see if one is a multiple of the other.
 
  • #8
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thanks tiny-tim

hehe sorry, im pretty bad at that HallsofIvy.
and ummm no?? i dont think so... :S
does that in itself tell us that it is not parallel?


so... it is correct the way i did it? even if it perhaps wasn't the simplest?
 

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