For what values of a are teh vectors parallel

[Painguy]

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.

 

[ehild]

For two vectors parallel, one is a scalar multiple of the other. w=λu.

Write it out in components.

ehild

 

[PhysicsandSuch]

If you can't use the dot product, how about the cross product instead?

 

[Painguy]

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.

 

[PhysicsandSuch]

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?

 

[Painguy]

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.

 

[PhysicsandSuch]

Sure thing!

 

[ehild]

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

 

[lendav_rott]

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}