How prove this $d(\alpha,W)$ is Gram determinant? if $W$ is subspace of $V$.and $V$ is $n$-dimensional Euclidean space. and $\alpha\in V$, Define the distance from  $\alpha$  to $W$

$$d(\alpha,W)=|\alpha-\alpha'|$$

where $\alpha'$ is $\alpha$  orthogonal projection on the  subspace  $W$.
such $\alpha_{1},\alpha_{2},\alpha_{3},\cdots,\alpha_{m}$ is $W$ one of  basis.
show that

$$d(\alpha,W)=\sqrt{\dfrac{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m},\alpha)|}{|G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})|}}$$
  where $G(\alpha_{1},\alpha_{2},\cdots,\alpha_{m})$ is Gram matrix.

my try: since this Gram determinant define:http://en.wikipedia.org/wiki/Gramian_matrix
But I can't prove this problem,I hope someone can help me.Thank you 
 A: If you have covered the relation of the determinant of a Gram matrix and the (hyper)volume of the $m$-dimensional parallelotope $V_m(\alpha_1,\ldots,\alpha_m)$ determined by having vectors $\alpha_1,\ldots,\alpha_m$ as its edges, namely
$$
V_m(\alpha_1,\ldots,\alpha_m)=\sqrt{|G(\alpha_1,\ldots,\alpha_m)|},
$$ 
then your claim follows easily. It is the $(m+1)$-dimensional generalization of the fact that the area of a parallelogram (in the plane) is its base times the height, where height equals the distance of the endpoint of the vector $\alpha_2$ from the baseline generated by $\alpha_1$. In the high-dimensional analogue the $m$-dimensional parallelotope with edges $\alpha_1,\ldots,\alpha_m$ takes the role of the "base", and the distance of the endpoint of the remaining edge vector $\alpha_{m+1}$ from the base takes the role of "height" $h$. So
from the equation 
$$
V_{m+1}(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})=V_m(\alpha_1,\ldots,\alpha_m)\cdot h
=V_m(\alpha_1,\ldots,\alpha_m)\cdot d(\alpha_{m+1},W)
$$
we get
$$
d(\alpha_{m+1},W)=h=\frac{V_{m+1}(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})}{V_m(\alpha_1,\ldots,\alpha_m)}=
\sqrt{\frac{|G(\alpha_1,\ldots,\alpha_m,\alpha_{m+1})|}{|G(\alpha_1,\ldots,\alpha_m)|}}.
$$
