I was reading AMD's answer here , and he considers three plane which represent three lines, now for three planes to intersect other than origin it must be that the all the normals together are not perpendicular, using this we can get the condition on the scalar triple product of coefficents i.e: if our lines are given as:

$$ a_1 x + b_1 y + c_1 =0 $$ $$ a_2 x + b_2 y + c_2 = 0$$ $$ a_3 x + b_3 y +c_3 = 0$$

Then, if:

$$ v= <a_1 , b_1 , c_1>$$ $$ u = <a_2,b_2,c_2>$$


$$ q=<a_3,b_3,c_3>$$

Must have scalar triple product as zero:

$$ q \cdot ( v \times u ) = 0$$

Is a 'required' condition for the lines to intersect, but is it sufficient for it to always gaurentee that it would? As in how do we know that when we put $z=1$ , that is where the planes intersect?


1 Answer 1


Yes it is a necessary and sufficient condition if you include as "point of intersection" the point at infinity in the case of parallel lines.

If the two lines $l_1: \; a_1x+b_1y+c_1=0,\quad l_2:\; a_2x+b_2y+c_2=0$ intersect in a point than $l_3=\lambda l_1 +\mu l_2$ also passes though that point. And the their $3 \times 3$ determinant is null.
Viceversa, if the determinant is null, then the rows are dependent and one can be expressed as a linear combination of the others.

-- reply to your comment --

In analytic or affine geometry, the introduction of the homogeneous coordinates render the panorama more ..homogeneous, in fact.
In particular we have the interesting duality points/lines in 2D, and points /planes in 3D lines.
in 2D, two lines always have a common point same as two points have a common line.
Translated to algebra, given $$l_1: \; a_1x_1+b_1x_2+c_1x_0=0,\quad l_2:\; a_2x_1+b_2x_2+c_2x_0=0$$ then $$l_1 \cap l_2$$ always has one non-trivial solution, either in terms of a common point $(x_1,x_2,x_0)$ in the above, and in terms of a line common to two points in $$a x_{1,1}1+bx_{1,2}+cx_{1,0}=0 \; \cap \; a x_{2,1}1+bx_{2,2}+cx_{2,0}=0$$

Then you know that in a homogeneous system prescribing the common point among three lines, a necessary and sufficient condition is that the determinant of the coefficients ($3 \times 3$) be null, because otherwise (rank of the matrix $3$) you have the only solution $(0,0,0)$ which does not correspond to a defined point.
Then if the rank of the matrix is $2$, you have a homogeneous solution $\lambda (\xi, \eta, \zeta)$ which is one point,
and if the rank is $1$ you get a homogeneous solution depending on two parameters, which is a line and which is the same as any of the three in the system.

The same, reverted, for three points sharing a common line.

  • $\begingroup$ I didn't get the inclusion point at infinity idea, it'd be nice if you could elaborate on it a bit more. It's pretty late at my place and I'm heading to sleep, so, I'll most prolly reply in about eight hours but again thank you for the answer ^^ $\endgroup$ Feb 6, 2021 at 23:47
  • $\begingroup$ it's late for me as well: will talk tomorrow $\endgroup$
    – G Cab
    Feb 6, 2021 at 23:54
  • $\begingroup$ @Buraian: added a concise hint about the introduction of homogeneous coordinates $\endgroup$
    – G Cab
    Feb 7, 2021 at 15:38
  • $\begingroup$ in 2D, two lines always have a common point same as two points have a common line. , I don't get this point you've written $\endgroup$ Feb 7, 2021 at 16:02
  • $\begingroup$ Please have a deep look into homogeneous coordinates, in the space available here I cannot go into single passages: just get convinced that in 2D a point is described by three coordinates $x_1,x_2,x_0$ with $x=x_1/x_0, y=x_2/x_0$ and where $(x_1,x_2,x_0)$ and $\lambda (x_1,x_2,x_0)$ represent the same geometric point. $\endgroup$
    – G Cab
    Feb 7, 2021 at 16:12

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