1. ## easy vector problem

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

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

2. Um... not quite.

The general formula:

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

Here, the angles are equal, so we get:

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

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

Equate cos theta;

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

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

The general formula:

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

Here, the angles are equal, so we get:

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

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

Equate cos theta;

$\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$
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 $\displaystyle 3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ for $\displaystyle x.$

7. Originally Posted by Plato
Just solve $\displaystyle 3\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ for $\displaystyle x.$
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?
$\displaystyle \left\| {\overrightarrow b } \right\| = \sqrt {10} \;\& \,\left\| {\overrightarrow c } \right\| = 3\sqrt {10}$
The 3 is left after division.
$\displaystyle \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|}$

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