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My algorithm to determine whether two lines are coincident (which have been proven previously they are parallel) verifies the following equation:

$$ \dfrac{x - x_o}{a} = \dfrac{y - y_o}{b} = \dfrac{z - z_o}{c} = \lambda $$

$x$, $y$ and $z$ are pointsB from lineB which I want to check in lineA
But if the director vector (not null) of the line contains some coordinate equal to zero $(a, b, c)$ my algorithm it will be invalid.
So how do I proceed in this case? I want to avoid using linear systems

Daniela Morais
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$$ \frac{x-x_0} a = \frac{y-y_0} b \tag 1 $$ is equivalent to $$ b(x-x_0) = a(y-y_0) \tag 2 $$ unless $a$ or $b$ is $0$. But you have another equality: $$ \frac{y-y_0} b = \frac{z-z_0} c \tag 3 $$ equivalent (unless $b$ or $c$ is $0$) to $$ c(y-y_0) = b(z-z_0). \tag 4 $$ Similarly we get $$ c(x-x_0) = a(z-z_0). \tag 5 $$ In cases where two of the three coefficients are $0$, one should check all three of $(2)$, $(4)$, and $(5)$.

Michael Hardy
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  • And if a and b are equal to zero and c nonzero? – Daniela Morais Aug 03 '15 at 19:30
  • You could have one or two of the three equal to $0$, but not all three, since then you don't have a direction vector. ${}\qquad{}$ – Michael Hardy Aug 03 '15 at 19:37
  • If two coordinates are equal to zero what I do with the coordinate that is not equal to zero? I equate it with what? I don't have the value of lambda – Daniela Morais Aug 03 '15 at 19:43
  • If $a=b=0$ and $c\ne0$ then $(4)$ would imply that $y=y_0$. In that case you'd need to check a third equality: $c(x-x_0) = a(z-z_0)$. That would imply $x=x_0$. The value of $z$ could then be anything at all, so the graph is just the set of points were $x$ and $y$ have the specified values. ${}\qquad{}$ – Michael Hardy Aug 03 '15 at 19:46