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Trying to solve for a common magnitude of two non-parallel vectors such that they intersect. I am currently solving using a left inverse, but I am not sure if this is correct.

Let $ x_1 = x_1^0 + tu_1\space$ and $y_1 = y_1^0 +t v_1$ be the parametric equations for the first vector, and $x_2 = x_2^0 + tu_2\space$ and $y_2 = y_2^0 + tv_2$ be the parametric equations for the second vector. To solve for a common factor, $t$, set both equations equal to each other:

$ \begin{bmatrix} x_1^0 \\ y_1^0 \end{bmatrix} + t\begin{bmatrix} u_1 \\ v_1 \end{bmatrix} = \begin{bmatrix} x_2^0 \\ y_2^0 \end{bmatrix} + t\begin{bmatrix} u_2 \\ v_2 \end{bmatrix} $

and solve for $t$

$t= \frac{\begin{bmatrix} x_1^0 - x_2^0 \\ y_1^0 - y_2^0 \end{bmatrix}}{\begin{bmatrix} u_2 - u_1 \\ v_2 - v_1 \end{bmatrix}} $

Of course no such operation exists for vectors, so setting $ \begin{bmatrix} x_1^0 - x_2^0 \\ y_1^0 - y_2^0 \end{bmatrix} = d\vec{x} \space $ and $\begin{bmatrix} u_2 - u_1 \\ v_2 - v_1 \end{bmatrix} = d\vec{u} $, I am taking the left inverse of $d\vec{u}$ to solve for $t$:

$t = (d\vec{u}^Td\vec{u})^{-1}d\vec{u}^Td\vec{x}$

I have tried visualising the results using MATLAB, but they are not consistent with my predictions. I feel as though I have made a mistake with my math. Any insight would be $\textit{greatly}$ appreciated.

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1 Answer 1

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At the point of intersection, the values of the parameters do NOT have to be the same. So the correct equation for the desired point of intersection is $$\begin{bmatrix} x_1^0 \\ y_1^0 \end{bmatrix} + t\begin{bmatrix} u_1 \\ v_1 \end{bmatrix} = \begin{bmatrix} x_2^0 \\ y_2^0 \end{bmatrix} + s\begin{bmatrix} u_2 \\ v_2 \end{bmatrix},$$ which you need to solve for two unknown parameters $t$ and $s$. Now this is just a system of two linear equations with two unknowns.

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  • $\begingroup$ I appreciate your comment, although I am trying to solve for a common factor in this case so I would like for both magnitudes to be equal. $\endgroup$
    – mykhaylo
    Commented Mar 26, 2020 at 19:27
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    $\begingroup$ @mykhaylo In general, the intersections of these lines do not occur at the same value of the parameter. Most of the time there will be no solution to your system of equations if you insist on this. $\endgroup$
    – amd
    Commented Mar 26, 2020 at 19:30
  • $\begingroup$ I see what you mean. It appears as though I am setting up the equation incorrectly. Solving for an intersection means that the points must be equal. I am looking for a value of t such that an intersection exists. $\endgroup$
    – mykhaylo
    Commented Mar 26, 2020 at 19:36
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    $\begingroup$ @mykhaylo: Then find the intersection first, and then check whether $s=t$ or not. $\endgroup$
    – zipirovich
    Commented Mar 26, 2020 at 19:54
  • $\begingroup$ Yes, and in the likely instance that they are not equal, I should be able to factor out the lowest common factor to get me the value I am looking for. $\endgroup$
    – mykhaylo
    Commented Mar 26, 2020 at 20:01

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