I know that this question had been asked before, but I am not exactly following what the answers say. Doing my own way here: I am puuzzled how to continue?
I named the points A,B,C, and the foot of the perpendiculars D and E. Now the point D would be like: $$d=kb+(1-k)c$$ Also it must satisfy $$(d-a)\cdot(c-b)=0$$ Now I use another set of vectors: $$\alpha=a-b,\beta=b-c,\gamma=c-a$$ So I end up with: $$d=b+\frac{\alpha.\beta}{\beta.\beta}\beta$$ Now for the orthocentre$(h)$ would be like: $$h=ma+(1-m)d=nc+(1-n)e\\m(a-d)-n(c-e)=e-d$$ Now how can I get $m,n$? and prove concurrence?
Linear equations
When you write $ma+(1−m)d=nc+(1−n)e$ or $m(a−d)−n(c−e)=e−d$ you have a linear system of equations: two equations (one for each coordinate of the resulting vector) in two variables ($m$ and $n$). You could write that in matrix and vector notation, as
$$\begin{pmatrix}a_1-d_1 & e_1-c_1\\a_2-d_2 & e_2-c_2\end{pmatrix} \cdot\begin{pmatrix}m\\n\end{pmatrix} =\begin{pmatrix}e_1-d_1\\e_2-d_2\end{pmatrix}$$
Solving that will usually break down to either Gaussian elimination or the computation of an inverse matrix. So I don't see how you could keep your vector notation without jumping through hoops for it.
Anternative: Descriptive not constructive
If your end goal is prooving that the altitudes intersect, you have another option: you can proove that there exists a point $h$ satisfying
$$(a-h)\cdot(b-c)=0\qquad (b-h)\cdot(c-a)=0\qquad (c-h)\cdot(a-b)=0$$
i.e. such that the line connecting $h$ to one of the corners is orthogonal to the edge opposite that corner. Reformulating the above you get
$$a\cdot(b-c)=h\cdot(b-c)\qquad b\cdot(c-a)=h\cdot(c-a)\qquad c\cdot(a-b)=h\cdot(a-b)$$
This, too, can be written as a system of linear equations:
$$ \begin{pmatrix} b_1-c_1 & b_2-c_2 \\ c_1-a_1 & c_2-a_2 \\ a_1-b_1 & a_2-b_2 \end{pmatrix} \cdot\begin{pmatrix} h_1 \\ h_2 \end{pmatrix} = \begin{pmatrix} a\cdot(b-c) \\ b\cdot(c-a) \\ c\cdot(a-b) \end{pmatrix} $$
In this setup, you'd have to proove that the system of linear equations still has a solution, although it is over-determined. You can prove this by showing that the third row is a linear combination of the first two rows. It's the same linear combination both on the left and on the right hand side of the equals sign. Therefore, any $h$ which satisfies the first two equations will satisfy the third as well.
Alternative: Projective Geometry
As an alternative, I'd tackle this using projective geometry and homogeneous coordinates. There, a point $(x,y)$ is represented by a vector $(x,y,1)^T$ or any multiple thereof. The line joining two points can be computed using the cross product, as can be the point of intersection between two lines. The matrix $D=\operatorname{diag}(1,1,0)$ multiplied with a line will yield a point at infinity orthogonal to that line. So you'd get
$$d=((D\cdot(b\times c))\times a)\times(b\times c)$$
as „join $b$ and $c$, take the point at infinity orthogonal to that, connect it with $a$ and intersect the result with the line joining $b$ and $c$.“ But you don't actually need the point $d$, you only need the altitude. So you'd have to show that the following three lines are concurrent:
$$ (D\cdot(b\times c))\times a\qquad (D\cdot(c\times a))\times b\qquad (D\cdot(a\times b))\times c $$
Three lines are concurrent if the determinant formed by their vectors is zero. So you could plug the above in a determinant, use variables as the coordinates of the vectors, and in the end obtain a big zero confirming your claim.
Of course, understanding why all this machinery works as claimed requires quite a bit of work, and exceeds the scope of this answer post. I just had the impression that you'd prefer to give closed form expressions for geometric objects, and projective geometry is great for this.