vector a = (2,5) and vector b = (-1,t)

solve for t when the vectors have a 45 degree angle between them

any help would be appreciated. :)

- Jan 11th 2011, 07:57 AMcoloradosolve for variable when vector a and b have a 45 degree angle between them
vector a = (2,5) and vector b = (-1,t)

solve for t when the vectors have a 45 degree angle between them

any help would be appreciated. :) - Jan 11th 2011, 08:08 AMe^(i*pi)
If I remember correctly you can use the dot product

$\displaystyle a.b = |a||b|cos \theta$

$\displaystyle -2+5t = \sqrt{29} \cdot \sqrt{1+t^2} \cdot \dfrac{\sqrt{2}}{2}$ - Jan 11th 2011, 08:27 AMcolorado
hm, yeah thats what I figured. I can't figure out how to solve from there on.

- Jan 11th 2011, 08:28 AMProve It
Where are you having trouble? It just ends up becoming a quadratic after you square both sides.

- Jan 11th 2011, 08:32 AMe^(i*pi)
$\displaystyle -2+5t = \dfrac{\sqrt{58(1+t^2)}}{2}$

$\displaystyle 10t-4 = \sqrt{58+58t^2}$

$\displaystyle 100t^2-80t+16 = 58+58t^2$

$\displaystyle 42t^2-80t-42 = 0$

$\displaystyle 21t^2-40t-21 = 0$

This one does factor but it might be easier to use the quadratic formula - Jan 11th 2011, 09:03 AMcolorado
I tried that but Ididn't get the correct answer. the answer is t=7/3

- Jan 11th 2011, 09:24 AMQuacky
Yes, that is the positive answer you get when you solve the quadratic.

$\displaystyle 21t^2 - 40 t - 21 = 0$

$\displaystyle t = \displaystyle\frac{40 \pm \sqrt{(-40)^ 2- 4(21)(-21)}}{2\times 21}$

$\displaystyle = \displaystyle\frac{40 \pm 58}{42}$ - Jan 11th 2011, 09:29 AMcolorado
ah okay, i'll give it another go

- Jan 11th 2011, 09:35 AMQuacky
I've editted the post. Where was your mistake? I assume it was probably just a sign error or something.

- Jan 11th 2011, 10:32 AMmr fantastic
- Jan 11th 2011, 11:42 AMcolorado
noted