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When doing the exercise of finding a line between two intersection of planes, we require to find a point and direction vector of the line. The direction vector is easy because it's perpendicular to both the normals, but I'm a tad-bit confused about how to take the point.

Suppose, we were to be given equation of two planes,

$$P_1 : A_1 x + B_1 y +C_1 z+ D = 0$$

And,

$$ P_2 : A_2 x +B_2 y +C_2 z +D = 0$$

To find a point along the line of intersection, it is often instructed to put one of the coordinates as zero, say $x, y$ or $z$ and then solve for the remaining coordinates. But, I'm not sure why we do this, as in , how do we know that the line between intersection of two lines would always need to have $x$ , $y$ and $z$ intercepts?

I saw this post but didn't think it addressed my query and neither was it addressed in this one

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  • $\begingroup$ It won't necessarily have $x,y$ and $z$ intercepts, but it will necessarily have an $x,y$ or $z$ intercept. $\endgroup$ Commented Aug 16, 2020 at 10:42
  • $\begingroup$ How would one prove this? $\endgroup$ Commented Aug 16, 2020 at 10:44
  • $\begingroup$ @DDD4C4U you wont need a rigourous proof for that, the line maybe parallel to a plane hence eliminating intercepts either on x,y or z $\endgroup$ Commented Aug 16, 2020 at 10:46
  • $\begingroup$ Do you have some knowledge about matrix calculus? Then it is not too difficult to explain what is going on. $\endgroup$ Commented Aug 16, 2020 at 10:48
  • $\begingroup$ Mhm yes, I think I know a thing or two $\endgroup$ Commented Aug 16, 2020 at 10:49

2 Answers 2

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Suppose that $\left|\begin{matrix} A_1&B_1 \\ A_2 & B_2\end{matrix}\right| = A_1B_2-B_1A_2\neq 0$. Then you may reformulate the problem as follows:

$$\begin{pmatrix} A_1&B_1 \\ A_2 & B_2\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = -\begin{pmatrix} C_1 z + D_1 \\ C_2 z+D_2\end{pmatrix} $$ and solve for $x$ and $y$: $$ \begin{pmatrix} x \\ y\end{pmatrix} = -\begin{pmatrix} C'_1 z + D'_1 \\ C'_2 z+D'_2\end{pmatrix} $$ This shows that for any $z=t\in{\Bbb R}$ you get a unique solution for $x$ and $y$. What happens here is that the intersection of the two planes $P_1,P_2$ with the plane $z-t=0$ provides two non-parallel lines (because of the non-zero A-B determinant) in the $x-y$ plane. These two lines therefore have a unique intersection point.

Now, when your A-B determinant above is zero (so your two lines in the $x-y$ plane are parallel) then you may look for a non-zero $B-C$ matrix (and solve for $y,z$) or a non-zero $C-A$ matrix (and solve for $z,x$). If all these determinants are zero then your two original planes are in fact parallel so either the intersection is empty or it is a plane.

Note that the three determinants you compute are in fact the component of the cross-product of normal vectors for the planes, so the cross-product being non-vanishing is indeed a condition for the intersection to be a line.

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One might solve the such questions either by assuming any of $(x,y,z)$ to be zero, or keeping one as a constant. The intuition behind keeping one of them zero is that, most of the times the lines we get are not parallel to a plane so they must definitely intersect.

When it's not such a case, keeping the variable zero would yield inconsistent pair of linear equations.

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