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    <p>Using the notation given in <a href="http://pomax.github.io/bezierinfo/#extremities" rel="nofollow">http://pomax.github.io/bezierinfo/#extremities</a>, a quadratic Bezier curve is given by:</p> <pre><code>B(t) = P1*(1-t)**2 + 2*P2*(1-t)*t + P3*t**2 </code></pre> <p>Therefore, (by taking the <a href="http://en.wikipedia.org/wiki/Derivative" rel="nofollow">derivative</a> of <code>B</code> with respect to <code>t</code>) the tangent to the curve is given by</p> <pre><code>B'(t) = -2*P1*(1-t) + 2*P2*(1-2*t) + 2*P3*t = 2*(P1 - 2*P2 + P3)*t + 2*(-P1 + P2) </code></pre> <p>Given some tangent direction <code>V</code>, solve </p> <pre><code>B'(t) = V </code></pre> <p>for <code>t</code>. If there is a solution, <code>t = ts</code>, then the point on the Bezier curve which has tangent direction <code>V</code> is given by <code>B(ts)</code>.</p> <hr> <p>We essentially want to know if two vectors (<code>B'(t)</code> and <code>V</code>) are parallel or anti-parallel. There is a trick to doing that.</p> <p>Two vectors, <code>X</code> and <code>Y</code>, are perpendicular if their <a href="http://en.wikipedia.org/wiki/Dot_product#Geometric_definition" rel="nofollow">dot product</a> is zero. If <code>X = (a,b)</code> and <code>Y = (c,d)</code> then the dot product of <code>X</code> and <code>Y</code> is given by</p> <pre><code>a*c + b*d </code></pre> <p>So, <code>X</code> and <code>Y</code> are parallel if <code>X</code> and <code>Y_perp</code> are perpendicular, where <code>Y_perp</code> is a vector perpendicular to <code>Y</code>.</p> <p>In 2-dimensions, if in coordinates <code>Y = (a,b)</code> then <code>Y_perp = (-b, a)</code>. (There are two perpendicular vectors possible, but this one will do.) Notice that -- using the formula above -- the dot product of <code>Y</code> and <code>Y_perp</code> is</p> <pre><code>a*(-b) + b*(a) = 0 </code></pre> <p>So indeed, this jibes with the claim that perpendicular vectors have a dot product equal to 0.</p> <p>Now, to solve our problem: Let</p> <pre><code>B'(t) = (a*t+b, c*t+d) V = (e, f) </code></pre> <p>Then <code>B'(t)</code> is parallel (or anti-parallel) to <code>V</code> if or when <code>B'(t)</code> is perpendicular to <code>V_perp</code>, which occurs when</p> <pre><code>dot product((a*t+b, c*t+d), (-f, e)) = 0 -(a*t+b)*f + (c*t+d)*e = 0 </code></pre> <p>We know <code>a</code>, <code>b</code>, <code>c</code>, <code>d</code>, <code>e</code> and <code>f</code>. Solve for <code>t</code>. If <code>t</code> lies between 0 and 1, then <code>B(t)</code> is part of the Bezier curve segment between <code>P1</code> and <code>P3</code>.</p>
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    1. POHow to find a point (if any) on quadratic Bezier with a given tangent direction?
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    1. POHow to find a point (if any) on quadratic Bezier with a given tangent direction?
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    1. COSorry, I forgot to specify that I need to compare normalized direction vectors since i'm comparing points from different curves... Can see how to rearrange `B'(t)=V` to get `t` but am stuck on how to extend this to solve `B'(t) / |B'(t)| = V / |V|`
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    2. COYou do not need to normalize the vectors to determine if they are parallel or anti-parallel. I've edited the post to show how. It's a long explanation, but in the end the computation is short.
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    3. COHey unutbu, just noticed while debugging my code that the definition the second term in your definition of B(t) should actually be 2 * P2 * (1-t)*t according to [wikipedia](https://en.wikipedia.org/wiki/B%C3%A9zier_curve#Quadratic_B.C3.A9zier_curves)
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