In the case below, calculate (i) the orthogonal projection
$ P_W $ associated with the two-dimensional vector subspace
W = L (S ') of the vector space V generated by
linearly independent set S '= $ \{\vec f_1, \vec f_2 \} $ applying orthonormalization from Gram-Schmidt to S'.
(ii) calculate the area of the orthogonal projection on W of the parallelogram generated by the vectors $\vec x_1, \vec x_2 \in V $ of two different ways: by the associated determinant function to the orthonormalization S = $ \{\vec e_1, \vec e_2 \} $ de S 'and at
determinant of the Gram matrix of $P_W \vec x_1, P_W \vec x_2 $
thus verifying Lagrange's identity.
V=$ R^3 $, S'=$\{\vec f_1 =(1,0,1),\vec f_2 = (1,-1,0)\},\vec x_1 =(1,1,1),\vec x_2= (1,0,0)$.
In (i)
$ \vec f_1 => \frac{\vec f_1}{|\vec f_1|} = (\frac{1}{\sqrt2},0\frac{1}{\sqrt2})= \vec v_1$
$\vec v_2 = \vec f_2 -\frac {\lt\vec f_2, \vec v_1 \gt}{\lt\vec v_1, \vec v_1 \gt}.\vec v_1 = (\frac{1}{2},-1, -\frac{1}{2})$
$\frac{\vec v_2}{|\vec v_2|}= (\frac{1}{\sqrt 6},\sqrt\frac{2}{3},-\frac{1}{\sqrt 6})$
It is correct?
So in (ii) I know that I need to calculate the area of the parallelogram in two different ways.
I think these two ways are:
A= det $\begin{pmatrix} i&j&k\\ a_1&a_2&a_3\\ b_1&b_2&b_3\\ \end{pmatrix}$and the area is given by
|A|= $\sqrt{ (a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2}$
In the second way the area would be given by
det $\begin{pmatrix} \langle a,a \rangle&\langle a,b \rangle\\ \langle b,a \rangle&\langle b,b \rangle\\ \end{pmatrix}$
It is correct so far?
Assuming that it is correct. Who are vectors a and b ?
Are they the orthogonal projection of $ x_1, x_2 $?
So do I do the same process done in (i) only this time in $ x_1, x_2 $?