# For what values of a are teh vectors parallel

• Painguy

#### Painguy

For what values of "a" are teh vectors parallel

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

## 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.

For two vectors parallel, one is a scalar multiple of the other. wu.

Write it out in components.

ehild

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

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.

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?

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.

Sure thing!

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

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}