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For what values of a are teh vectors parallel

  • Thread starter Painguy
  • Start date
  • #1
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For what values of "a" are teh vectors parallel

Homework Statement


v = 5a i -3 j
w = a^2 i -6 j

Homework Equations



The Attempt at a Solution



My first thought was too get the unit vectors of both and set them equal to each other and solve for a, but that got messy.

(5a i- 3j)/sqrt(25a^2 +9) = (a^2i -6j)/sqrt(a^4 +36)

I tried solving for a separately and got a=5, but its obvious that is wrong because i end up with
25 i -3 j =25 i -6 j

I'm not supposed to use the dot product for this problem.
 

Answers and Replies

  • #2
ehild
Homework Helper
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For two vectors parallel, one is a scalar multiple of the other. wu.

Write it out in components.

ehild
 
  • #3
If you can't use the dot product, how about the cross product instead?
 
  • #4
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Oh right. That seems to work out. So a=-10. The two also seem to be parallel if a=0 since you would end up with
-3j=6j
but that isn't something that results from doing what you told me. Is that a trivial case i should just notice or is there an algebraic way of solving that? Thanks for your help thus far.

I can't use the cross product either.
 
  • #5
Using the proportional method suggested by ehild, you do get both answers analytically:

If you do the math right, after solving for λ and substituting, you should end up with an equation that looks like $$ a^2 = 10 a$$

This has solutions ## a={10, 0}##. Perhaps you divided out the ##a## earlier on before you noticed the zero solution, which prevents that division?
 
  • #6
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Using the proportional method suggested by ehild, you do get both answers analytically:

If you do the math right, after solving for λ and substituting, you should end up with an equation that looks like $$ a^2 = 10 a$$

This has solutions ## a={10, 0}##. Perhaps you divided out the ##a## earlier on before you noticed the zero solution, which prevents that division?
Oh I see. Yeah I did do that division a bit earlier. That makes much more sense. Thanks very much for your help guys.
 
  • #8
ehild
Homework Helper
15,478
1,854
Oh right. That seems to work out. So a=-10.
No, that is wrong. Check.

The two also seem to be parallel if a=0 since you would end up with
-3j=6j
but that isn't something that results from doing what you told me. Is that a trivial case i should just notice or is there an algebraic way of solving that?

Both solution result from the method I suggested.

ehild
 
  • #9
223
10
The vectors are conviniently in a 2D world so you can work with slopes - assume any of the i or j unit vectors as x or y or what notation you might prefer. Tan(slope angle) is the vector's ordinate/abscissa so you get 5a/-3 = a^2/-6 or reverse the denoms and numerators since it's called the main property of an equality or something where a/b = c/d is the same as b/a = d/c. The slopes are equal because the vectors are parallel and you arrive at the same solution a = {0, 10}
 

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