# find t when vectors are parallel

• January 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
• January 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?
• January 10th 2011, 02:44 PM
pickslides
Find where the cross product of a and b is 0.
• January 10th 2011, 02:49 PM
could you show me how to solve for t in this problem?
• January 10th 2011, 02:52 PM
Plato
Can you solve $\dfrac{-1}{2}=\dfrac{t}{5}$ for $t~?$
• January 10th 2011, 02:53 PM
-5/2 =t?
• January 10th 2011, 02:55 PM
Plato
You have it!
• January 10th 2011, 02:55 PM
could you show me a way to solve for t using dot product?
• January 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 $\dfrac{-1}{2}=\dfrac{t}{5} .$
• January 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.
• January 10th 2011, 03:24 PM
pickslides
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

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