# find t when vectors are parallel

• Jan 10th 2011, 02:37 PM
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
• Jan 10th 2011, 02:41 PM
snowtea
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?
• Jan 10th 2011, 02:44 PM
pickslides
Find where the cross product of a and b is 0.
• Jan 10th 2011, 02:49 PM
could you show me how to solve for t in this problem?
• Jan 10th 2011, 02:52 PM
Plato
Can you solve $\displaystyle \dfrac{-1}{2}=\dfrac{t}{5}$ for $\displaystyle t~?$
• Jan 10th 2011, 02:53 PM
-5/2 =t?
• Jan 10th 2011, 02:55 PM
Plato
You have it!
• Jan 10th 2011, 02:55 PM
could you show me a way to solve for t using dot product?
• Jan 10th 2011, 03:13 PM
Plato
Quote:

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 $\displaystyle \dfrac{-1}{2}=\dfrac{t}{5} .$
• Jan 10th 2011, 03:17 PM
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.
• Jan 10th 2011, 03:24 PM
pickslides
Plato has the easiest way here.

I think you want to use the following

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

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

Solving this you should get

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