I am a second year open university undergraduate math student and I am taking a first course on vector calculus. I am trying to prove the below result in geometry using vector algebra. I would like to have my proof verified/simplified if necessary, any inputs will be extremely valuable.
As an aside, what's a good physics book for a typical undergraduate student, that applies the ideas in Vector calculus, that I can pleasure read and solve along side the mathematics?
Lines drawn through a point $E$ and the vertices $A,B,C,D$ of a tetrahedron cut the planes of the opposite faces in $A',B',C',D'$. Show that the sum of the ratios in which these points divide $EA,EB,EC,ED$ is $-1$.
Proof.
From the theorem on co-planar points, we know that, we can always assign non-zero weights $\alpha,\beta,\gamma,\delta$ to four co-planar points $A,B,C,D$, such that they form a self-centroidal set.
$\alpha A + \beta B +\gamma C + \delta D = 0\\ \alpha + \beta + \gamma + \delta = 0$
Consider the points $A',B',C',D'$ lying in the plane of the four triangular faces - $BCD$, $CDA$, $DAB$, $ABC$.
Since, $B,C,D,A'$ are co-planar, four non-zero weights can be assigned such that $B,C,D,A'$ form a self-centroidal set.
$\beta b + \gamma c + \delta d + \alpha' a' = 0\\ \beta + \gamma + \delta + \alpha' = 0$
$\implies\beta b + \gamma c + \delta d = (\beta + \gamma + \delta)a' $
On similar lines, we can write 4 equations for position vectors $a',b',c',d'$:
$\beta b + \gamma c + \delta d = (\beta + \gamma + \delta) a'$
$\alpha a + \gamma c + \delta d = (\alpha + \gamma + \delta) b'$
$\alpha a + \beta b + \delta d = (\alpha + \beta + \delta) c'$
$\alpha a + \beta b + \gamma c = (\alpha + \beta + \gamma) d'$
Subtracting expression (1) - (2),
$\beta b - \alpha a = (\beta + \gamma + \delta) a' - (\alpha + \gamma + \delta) b'\\ \beta b + (\alpha + \gamma + \delta) b' = \alpha a + (\beta + \gamma + \delta) a'\\ \frac{\beta b + (\alpha + \gamma + \delta) b'}{\alpha + \beta + \gamma + \delta} = \frac{\alpha a + (\beta + \gamma + \delta) a'}{\alpha + \beta + \gamma + \delta} = e$
$E$ is the point of intersection of $AA,BB'$. Moreover,
$\alpha a + (\beta + \gamma + \delta) a' = (\alpha + \beta + \gamma + \delta)e\\ a'=\frac{(\alpha + \beta + \gamma + \delta)e - \alpha a}{\beta + \gamma + \delta}$
Thus, $A'$ divides $EA$ externally in the ratio $-\alpha/(\alpha + \beta + \gamma + \delta)$. Likewise, $B',C',D'$ divide $EB,EC,ED$ externally in the ratios $-\beta/(\alpha + \beta + \gamma + \delta), -\gamma/(\alpha + \beta + \gamma + \delta), -\delta/(\alpha + \beta + \gamma + \delta)$ respectively. The sum of these ratios is :
$$\begin{aligned}&\frac{-\alpha}{\alpha + \beta + \gamma + \delta}+\frac{-\beta}{\alpha + \beta + \gamma + \delta}+\frac{-\gamma}{\alpha + \beta + \gamma + \delta}+\frac{-\delta}{\alpha + \beta + \gamma + \delta}\\&=-\frac{(\alpha + \beta + \gamma + \delta)}{\alpha + \beta + \gamma + \delta} \\&= -1 \end{aligned}$$