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I have two parallel line segments, say AB, CD. If I project the end points onto a common third parallel line, then I want to know the portion of overlap made by above 2 lines. I think I should use scalar of lines but, I am confused to figure out how to calculate this (I have 4 scalars).. I feel, when the line segments orient in opposite directions, then scalar might be +,-. then the case of so difficult for me to understand.

If anyone can, help me to figure out a procedure to do this.

thanks in advance.

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I assume that you are talking about line segments on a 2D plane and "projection" means "orthogonal projection". That is, the projection of $A$ on a line is the foot of perpendicular from $A$ to that line, and so on and so forth.

Let $A=(a_1,a_2),\ B=(b_1,b_2),\ C=(c_1,c_2)$ and $D=(d_1,d_2)$. Let the third line passes through some two points $P=(p_1,p_2)$ and $Q=(q_1,q_2)$. So the vector $$ \vec{u}=(u_1,u_2)=\frac{\vec{PQ}}{\|\vec{PQ}\|}= \frac{1}{\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}}(q_1-p_1,\,q_2-p_2). $$ has length $1$ and it is parallel to the third line. Now, if you set $P=(p_1,p_2)$ as the origin, and let the half of the third line that extends from $P$ and points to the direction of $\vec{u}$ as the "positive $u$-axis", and the other half as the "negative $u$-axis", then the "$u$-coordinate" of the projection of $A$ on the third line is given by $\vec{PA}\cdot\vec{u}$, i.e. $$(a_1-p_1)u_1+(a_2-p_2)u_2.$$ Replace $(a_1,a_2)$ in last expression by $(b_1,b_2), (c_1,c_2)$ or $(d_1,d_2)$, you get the $u$-coordinates of $B,C$ or $D$. You can now do your computation as if $A,B,C,D$ all lie on the same $u$-axis, with their corresponding $u$-coordinates.

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  • $\begingroup$ thank you very very much. but it is my mistake, i forgot to tell that I am having 3D lines. then, how would it be? same? $\endgroup$ – gnp Nov 27 '12 at 20:46
  • $\begingroup$ It depends on your definition of "projection". If you are talking about foot of perpendicular, then the analogous procedure will still give you the $u$-coordinates. However, since each point is now specified by 3 coordinates (x,y,z) instead of 2 (x and y), you must take care of the extra one. For instance, the dot product $\vec{PA}\cdot\vec{u}$ would become something like $(a_1-p_1)u_1+(a_2-p_2)u_2+(a_3-p_3)u_3$, etc.. Also, if the direction of your "projection" is not perpendicular to the third line, the story will be different. $\endgroup$ – user1551 Nov 27 '12 at 21:21

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