1. ## easy vector problem

Hi! Given $\vec a=<2x,1,1-x>$, $\vec b=<-1,3,0>$ and $\vec c=<5,-1,8>$ as three vectors, find x so that $\vec a$ forms equal angles with the vectors $\vec b$ and $\vec c$.

I tried this $a \cdot b = a \cdot c$ ( $\cdot$ is a dot product). Then I isolated $cos\alpha$ and $cos\beta$ and equated them. Then I solved by x. Is this correct?

2. Um... not quite.

The general formula:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

Here, the angles are equal, so we get:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

$\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos\theta$

Equate cos theta;

$\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$

3. Originally Posted by Unknown008
Um... not quite.

The general formula:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

Here, the angles are equal, so we get:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta$

$\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos\theta$

Equate cos theta;

$\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$
That's what I was talking about. What wasn't right then?

4. From what I understood, you didn't divide by the product of the magnitudes of the vectors.

5. I did. I wasn't clear, my fault. Thank you very much.

6. Just solve $3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ for $x.$

7. Originally Posted by Plato
Just solve $3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ for $x.$
Could you explain why you used 3a?

I admit that solving from the formula isn't that easy with the quadratic equations

8. Originally Posted by Unknown008
Could you explain why you used 3a?
$\left\| {\overrightarrow b } \right\| = \sqrt {10} \;\& \,\left\| {\overrightarrow c } \right\| = 3\sqrt {10}$
The 3 is left after division.
$\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$

9. Oh gosh, right, I don't know how I missed that part >.< Thanks for that