# Thread: find t when vectors are parallel

1. ## find t when vectors are parallel

I have this question vector a = (2,5) vector b = (-1,t)
find t when a and b are parallel.

I can't seem to figure this one out. Theres nothing in my text about how to solve this. any help is appreciated

2. The two vectors are parallel when (2,5) = a(-1,t) = (-a, at) where a is some constant.
i.e. parallel vectors are vectors that differ only in magnitude but not in direction.

So now you have:
-a = 2
at = 5

Can you solve for t?

3. Find where the cross product of a and b is 0.

4. could you show me how to solve for t in this problem?

5. Can you solve $\dfrac{-1}{2}=\dfrac{t}{5}$ for $t~?$

6. -5/2 =t?

7. You have it!

8. could you show me a way to solve for t using dot product?

could you show me a way to solve for t using dot product?
Actually the notation of parallel vectors is not usually associated with dot product.
The dot product of two perpendicular vectors is zero.

Rather parallel vectors are scalar multiples of each other.
Therefore, their components are proportional.
That is why we have $\dfrac{-1}{2}=\dfrac{t}{5} .$

10. thanks, yeah that occurred to me as a walked away to take a break. thanks a bunch. this forum has been a life saver already.

11. Plato has the easiest way here.

I think you want to use the following

$\displaystyle \cos \theta = \frac{a\cdot b}{\|a\|\|b\|}$

$\displaystyle \cos 0 = \frac{2\times -1 +5\times t}{\sqrt{29}\sqrt{1+t^2}}$

Solving this you should get

$\displaystyle (2t+5)^2=0\implies t = \frac{-5}{2}$