# How to find a unit vector parallel to vector which related to other two in a cube?

#### Chemist116

The problem is as follows:

The figure from below shows a cube of side $a$ where $C$ and $D$ are midpoints of the edges. Find a unit vector paralell to vector $\vec{x}$ such that $\overrightarrow{AC}\cdot \vec{x}=0$ and $\overrightarrow{BD}\cdot\vec{x}=0$

The alternatives are as follows:

$\begin{array}{ll} 1.&\left(\frac{\sqrt{17}}{17}\right)(\hat{j}+4\hat{k})\\ 2.&\left(\frac{\sqrt{2}}{2}\right)(-\hat{j}+\hat{k})\\ 3.&\left(\frac{\sqrt{5}}{5}\right)(\hat{j}+2\hat{k})\\ 4.&\left(\frac{\sqrt{10}}{10}\right)(3\hat{j}+\hat{k})\\ 5.&(\frac{\sqrt{26}}{26})(\hat{j}+5\hat{k})\\ \end{array}$

This problem has left me to go in circles. The only thing which I was able to establish in this cube was that the vectors

$\overrightarrow{BD}=a(\left \langle +1,+1, -\frac{1}{2} \right \rangle)$

$\overrightarrow{AC}=a(\left \langle -1,+1, -\frac{1}{2} \right \rangle)$

However this is the part where I got stuck. How exactly can I find the vector which is requested?. Help! please.

#### romsek

Math Team
This problem is terribly worded.

so you have $A = (a,0,a),~B = (0,0,a),~C=(0,a,a/2),~D=(a,a,a/2)$

To find $\vec{x}$ that is orthogonal to both $\vec{AC}$ and $\vec{BD}$ we find $\vec{AC}\times \vec{BC}$

$\vec{AC} = C-A = (-a,a,-a/2)$
$\vec{BD} = D-B = (a,a,-a/2)$

$u=\vec{AC}\times \vec{BD} = a^2 (0,-1,-2)$

$\hat{u} = \dfrac{u}{\|u\|} = \dfrac{1}{\sqrt{5}}(0,-1,-2) = -\dfrac{1}{\sqrt{5}}j-\dfrac{2}{\sqrt{5}}k$

Now we have to translate this into one of your options.
A vector parallel to this vector will simply be the negative of it, and we can rewrite it in normalized form.

$-\hat{u} = \dfrac{\sqrt{5}}{5} j + \dfrac{2\sqrt{5}}{5}k = \dfrac{\sqrt{5}}{5}(j + 2k)$

i.e. option 3