# Thread: solve for variable when vector a and b have a 45 degree angle between them

1. ## solve 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.

2. If I remember correctly you can use the dot product

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

$-2+5t = \sqrt{29} \cdot \sqrt{1+t^2} \cdot \dfrac{\sqrt{2}}{2}$

3. hm, yeah thats what I figured. I can't figure out how to solve from there on.

4. Where are you having trouble? It just ends up becoming a quadratic after you square both sides.

5. $-2+5t = \dfrac{\sqrt{58(1+t^2)}}{2}$

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

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

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

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

This one does factor but it might be easier to use the quadratic formula

6. I tried that but Ididn't get the correct answer. the answer is t=7/3

7. Yes, that is the positive answer you get when you solve the quadratic.

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

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

$= \displaystyle\frac{40 \pm 58}{42}$

8. ah okay, i'll give it another go

9. I've editted the post. Where was your mistake? I assume it was probably just a sign error or something.

10. Originally Posted by colorado
I tried that but Ididn't get the correct answer. the answer is t=7/3
Please show your working rather than just giving an answer. Then it is easier to diagnose what help you actually need.

11. noted