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:

$a \cdot b = ab cos\theta$

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

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

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

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

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

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

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

$2250t^2 + 2250 = t^2$

$2249t^2 = 2250$

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

$t=1.00$

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

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:

$a \cdot b = ab cos\theta$

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

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

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

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

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

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

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

$2250t^2 + 2250 = t^2$

$2249t^2 = 2250$

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

$t=1.00$

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

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!!