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

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- Jan 10th 2011, 02:37 PMcoloradofind 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 PMsnowtea
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 PMpickslides
Find where the cross product of a and b is 0.

- Jan 10th 2011, 02:49 PMcolorado
could you show me how to solve for t in this problem?

- Jan 10th 2011, 02:52 PMPlato
Can you solve $\displaystyle \dfrac{-1}{2}=\dfrac{t}{5}$ for $\displaystyle t~?$

- Jan 10th 2011, 02:53 PMcolorado
-5/2 =t?

- Jan 10th 2011, 02:55 PMPlato
You have it!

- Jan 10th 2011, 02:55 PMcolorado
could you show me a way to solve for t using dot product?

- Jan 10th 2011, 03:13 PMPlato
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 PMcolorado
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 PMpickslides
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

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