# equation in matrix form

i have an equation

$$\sum_{i=0}^m [w^{(i)}[(c-a^{(i)})-\frac{d^{(i)}(c-a^{(i)})\cdot d^{(i)}}{dis^2}]]=0$$ Where:

$a,b$ are two end points in a 3D Line. $d$ is a vector and vector $d=b-a$. the magnitude of the vector $|d|=dis$. I have many lines and therefore, i have many $a,b$ points. thats why above equation represent $a,d$ as $a^{(i)},d^{(i)}$ . However, this $dis$ is single scalar value and others have $3$ value components. this $\cdot$ is the dot product of the vector $c-a$ and vector $d$. $w^{(i)}$ is the weight (of a line).

As i said earlier, I have many points which represent different lines and thus i am using $a^{(i)}$ and $d^{(i)}$ to represent a point on a line and direction vector of line. but I have a one point which is outside to the lines and represent as $c$. Assume $i$ is lasting from $0$ to $m$. If I say the coordinates of points $a^{(i)},b^{(i)}$ are as $(x_1,y_1,z_1)^{(i)},(x_2,y_2,z_2){(i)}$ respectively. and the coordinate of point $c$ as $(x_0,y_0,z_0)$. then How do I convert this equation in to matrix form $[A][X] = [B]$. where $X$ represent the componenets of $c$ as this $(x_0,y_0,z_0)$ are unknown that i need to solve.

Please convert this equation to matrix form!

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Please use latex to make your question more friendly. – Shiyu Sep 6 '11 at 8:02
@anon: i have modified the post. may be now it is clear. c is a point in 3D. – niro Sep 6 '11 at 8:48

Write $\hat{d}{}^i$ for $d^i/|d^i|$. Thus we can remove the factor of $dis^2$ from the original equation by replacing each $d^i$ with $\hat{d}{}^i$. Now let $\otimes$ denote the outer product. Furthermore note that the vector $\hat{d}{}^i(\hat{d}{}^i\cdot\vec{c})$ can actually be rewritten as the product $(\hat{d}{}^i\otimes\hat{d}{}^i)\vec{c}$, and similarly for the $a^i$s - I recommend checking this yourself for good exercise. Now we can define the matrix $P=\sum_{i=0}^mw^i(I-\hat{d}{}^i\otimes\hat{d}{}^i)$ and the vector $\vec{\alpha}=\sum_{i=0}^m\vec{a}{}^i$. Let $I$ be the identity matrix. Then the original equation can be abridged to just $P(\vec{c}-\vec{\alpha})=\vec{0}$. You write $\vec{x}=\vec{c}$ and desire the equation $A\vec{x}=\vec{b}$. You should be able to see what you want to do from there.

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sorry i couldnt get you. actually in my post i left weights, though it was in my equation, for the simplicity. but, i couldnt figure out my equation with the weights according to your way as i cant get it clearly. could you please elaborate how my equation simplify in the form of Ax=B. finally i want to solve values for (xo,yo,zo) which is in 'c' in my equation. So, i am looking for the equation as (AX=B) with all xs,ys & zs as I need this for my normal equation in least square solutions. Please help me. – niro Sep 6 '11 at 10:16
@g_niro: I've augmented the equation to involve the weights. – anon Sep 6 '11 at 10:52
thank you very much for the explanation u gave me. – niro Sep 10 '11 at 14:55

What anon is suggesting is pictorially this: $$\sum_{i=1}^n w_i\left(\begin{bmatrix} | \\ (c-a_i)\\ | \end{bmatrix} - \begin{bmatrix} | \\ \bar{d}_i\\ | \end{bmatrix}\begin{bmatrix} - &d_i^T&- \end{bmatrix} \begin{bmatrix} | \\ (c-a_i)\\ | \end{bmatrix} \right)=0$$ where $\bar{d}_i$ is the normalized $d_i$ (and dot product is commutative). Factoring out $c-a_i$ gives $$\sum_{i=1}^n w_i \underbrace{\left( I - \begin{bmatrix} | \\ \bar{d}_i\\ | \end{bmatrix}\begin{bmatrix} - &d_i^T&- \end{bmatrix} \right)}_{P_i}\begin{bmatrix} | \\ (c-a_i)\\ | \end{bmatrix}=0$$ You can take the $a_i$ terms to the other side. The remaining step is to find a way to include the sum into a matrix format: $$\underbrace{\begin{bmatrix} w_1I &w_2I &\dots w_nI \end{bmatrix} \begin{bmatrix} P_1\\P_2\\ \vdots \\P_n \end{bmatrix}}_{A}\underbrace{\begin{bmatrix}|\\c\\|\end{bmatrix}}_{x} = \underbrace{\begin{bmatrix} w_1I &w_2I &\dots w_nI \end{bmatrix} \begin{bmatrix} P_1 & & \\ &P_2\\ &&\ddots \\&&&P_n \end{bmatrix} \begin{bmatrix} a_1\\a_2\\ \vdots \\a_n \end{bmatrix}}_{b}$$ which is the $Ax=b$ form we wanted. Computationally, the form above is simply, ...ehm..., ridiculous. I just wanted to show you the general structure.

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thanks for the response. what i am trying is to find the point which is closest to 3 lines. after summing all 3 square distances (from that point to the lines) & getting derivatives with respect to $c$, then the equation should be arranged as $Ax=B$. So, in my equation $c=(x0,y0,z0)$ is the unknown and need to solve. since i have 3 lines, i have 3 values for $w$. it is really hard for me to figure out this equation with x,y,z coordniates as the coordinates have 3 components. any help please. – niro Sep 7 '11 at 0:22
though, i have 3 lines, i guess the demension of my matrics i.e. $A$ as 3x3, $x$ as 3x1 and $B$ as 3x1. (not sure) so, please show me how my equation would be, when we substitute $x,y,z$ & $w$ values. – niro Sep 7 '11 at 0:24
yes but this is already in the form that you want. I tried to sketch the dimensions with the $|$ and $-$ symbols. In other words, $c$ is the vector that you want to obtain. Check the labels of the underbraces, it is exactly in the form you wish to get. – user13838 Sep 7 '11 at 0:25
Let's go slowly. What is your first difficulty when reading my answer? – user13838 Sep 7 '11 at 0:26
what is the demension of $P1$? then what would be $w1I$, and $[w 1 I w 2 I …w n I ]$. if i have 4 lines, then what would be the dimension of matrix $A$. do we have 12 equations? if, possible please figure out this with x,y,z. plz. – niro Sep 7 '11 at 0:50