# Thread: quick vector question

1. ## quick vector question

Hi
I need help on the following:

Find the value of t that makes the angle between two vectors a=(3,1,0) and b(t,0,1) equal to 45 degrees.

This is what i have done:

$\displaystyle a \cdot b = ab cos\theta$

$\displaystyle \frac{a \cdot b}{ab} = 45$

$\displaystyle \frac{3t}{\sqrt{10}\sqrt{t^2+1}} = 45$

$\displaystyle 45\sqrt{10(t^2+1)} = 3t$

$\displaystyle \sqrt{10t^2 + 10} = \frac{3t}{45}$

$\displaystyle 10t^2 + 10 = \frac{t^2}{15^2}$

$\displaystyle 10t^2 + 10 = \frac{t^2}{225}$

$\displaystyle 225(10t^2 + 10) = t^2$

$\displaystyle 2250t^2 + 2250 = t^2$

$\displaystyle 2249t^2 = 2250$

$\displaystyle t=\sqrt{\frac{2250}{2249}}$

$\displaystyle t=1.00$

However the answer is$\displaystyle \frac{\sqrt{5}}{2}$.

P.S

2. Originally Posted by Paymemoney
Hi
I need help on the following:

Find the value of t that makes the angle between two vectors a=(3,1,0) and b(t,0,1) equal to 45 degrees.

This is what i have done:

$\displaystyle a \cdot b = ab cos\theta$

$\displaystyle \color{red}{\frac{a \cdot b}{ab} = 45}$

$\displaystyle \frac{3t}{\sqrt{10}\sqrt{t^2+1}} = 45$

$\displaystyle 45\sqrt{10(t^2+1)} = 3t$

$\displaystyle \sqrt{10t^2 + 10} = \frac{3t}{45}$

$\displaystyle 10t^2 + 10 = \frac{t^2}{15^2}$

$\displaystyle 10t^2 + 10 = \frac{t^2}{225}$

$\displaystyle 225(10t^2 + 10) = t^2$

$\displaystyle 2250t^2 + 2250 = t^2$

$\displaystyle 2249t^2 = 2250$

$\displaystyle t=\sqrt{\frac{2250}{2249}}$

$\displaystyle t=1.00$

However the answer is$\displaystyle \frac{\sqrt{5}}{2}$.

P.S
Dear Paymemoney,

There is a mistake in the second line. Can you find it??

3. is it suppose to be cos(45)??

4. found out my problem.

5. Originally Posted by Paymemoney
is it suppose to be cos(45)??
Dear Paymemoney,

Correct!!