# Thread: Reciprocal vectors

1. ## Reciprocal vectors

Q: Show that the vectors a=(1,2,1); b=(0,0,1) and c=(2,-1,1) for a non-orthogonal basis (I guess I can just show that a.b, a.c and b.c all doesn't equal 0?). Using the scalar triple product, write the vector d=(1,1,1) in terms of this basis.

I have been able to find the reciprocal vectors for a, b and c:
$\displaystyle \begin{array}{l} a' = \frac{1}{5}(i + 2j) \\ b' = \frac{1}{5}( - 3i - j + 5k) \\ c' = \frac{1}{5}(2i - j) \\ \end{array}$

but now I don't know what to do to re-write d in terms of the original 2 basis a, b, c. Help would be greatly appreciated.

2. Originally Posted by free_to_fly
Q: Show that the vectors a=(1,2,1); b=(0,0,1) and c=(2,-1,1) for a non-orthogonal basis ...

but now I don't know what to do to re-write d in terms of the original 2 basis a, b, c. Help would be greatly appreciated.
You have to show that there exist a triple of numbers (k, m , n) so that

$\displaystyle d = k \cdot a + m \cdot b + n \cdot c$ That means:

$\displaystyle (1,1,1) = k \cdot (1,2,1) + m \cdot (0,0,1) + n \cdot (2,-1,1)$

This yields a system of simultaneous equations:

$\displaystyle \left|\begin{array}{ccccccc}k& & & + & 2n & = & 1 \\ 2k& & & -&n&=&1 \\ k&+&m&+&n&=&1\end{array}\right.$

I've got $\displaystyle \left(k,m,n\right) = \left(\dfrac35\ ,\ \dfrac15\ ,\ \dfrac15\right)$

So $\displaystyle d = \left(\dfrac35\ ,\ \dfrac15\ ,\ \dfrac15\right)$ with respect to the base (a, b, c).