4
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For example if we define parametric line segment

curve1[t_, p1_, p2_] := t*p2 + (1 - t)*p1;

where p1 and p2 are 2D vectors, we can draw it easily

p1 = {0, 0}; p2 = {1, 2}; ParametricPlot[curve1[t, p1, p2], {t, 0, 1}]

enter image description here

But can't solve in general form

Clear[p1, p2]
Solve[curve1[t, p1, p2] == curve1[u, q1, q2], {t, u}]

because Mathematica treats p1, p2, q1 and q2 as scalars, and gives

$\left\{\left\{u\to -\frac{t (\text{p2}-\text{p1})}{\text{q1}-\text{q2}}-\frac{\text{p1}-\text{q1}}{\text{q1}-\text{q2}}\right\}\right\}$

Can I treat them as vectors and solve analytically?

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3
  • $\begingroup$ @DanielHuber of course not, if p1, p2, q1 and q2 are all different, it doesn't true. $\endgroup$
    – Dims
    Jan 13 at 10:38
  • $\begingroup$ Sorry I read it too fast. $\endgroup$ Jan 13 at 10:51
  • 1
    $\begingroup$ This may be a starting point: Reduce[curve1[t, p1, p2] == curve1[u, q1, q2] && (p1 | p2 | q1 | q2) ∈ Vectors[2, Reals], {t, u}] $\endgroup$
    – qreus
    Jan 13 at 10:54

2 Answers 2

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You can use Vectors.

Solve[curve1[t, p1, p2] == curve1[u, q1, q2] && (p1 | p2 | q1 | q2) ∈ Vectors[2, Reals] && (t|u) ∈ Reals, {t, u}] //FullSimplify

output

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3
$\begingroup$
Clear[p1, p2]
curve1[t_, p1_, p2_] := t*p2 + (1 - t)*p1;
{p1, p2, q1, q2} = {{p1x, p1y}, {p2x, p2y}, {q1x, q1y}, {q2x, q2y}};
Solve[curve1[t, p1, p2] == curve1[u, q1, q2], {t, u}]

enter image description here

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2
  • $\begingroup$ It's cheating :) Can I just tell solver, that some variables are vectors? $\endgroup$
    – Dims
    Jan 13 at 10:56
  • $\begingroup$ How do you know that it can be written without respect to the vector components? $\endgroup$ Jan 13 at 10:58

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