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Prove that the line of intersection of the planes $7x-4y+7z+16=0$, $4x+3y-2z+3=0$ is coplanar with the line of intersection of the planes $x-3y+4z+6=0$, $x-y+z+1=0$.

I can prove it by first converting both lines into symmetric form $\frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n}$ and using the condition of being coplanar.

Or I can just convert one of the lines in this form and write the equation of a general plane through the other one, then find the one plane which is parallel to the other line. If that plane contains the other line, both the lines are coplanar.

But both these methods requires us to first convert the equations of the lines into symmetric form.

Is there any method which can directly manipulate these four equations and identify whether they represent coplanar lines?

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2 Answers 2

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Hint:

You also simply can check the matrix with the coefficients of the planes as rows $$\left[\begin{array}{rrrr} 7&-4&7&16\\ 4&3&-2&3\\ 1&-3&4&6\\ 1&-1&1&1 \end{array}\right]$$ has rank $<4$, i.e. its determinant is $0$.

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  • $\begingroup$ This is equivalent to checking if the system of 4 linear equations is consistent. If $rank < 4$, we are guaranteed to have a solution, unique or infinite. In case of unique solution, the lines will intersect on a point, as the four plane are intersecting in a point. If the system has infinite solution, both lines must coincide. But this leaves an important case - lines can be parallel without intersecting and coinciding. In this case too there can be a plane passing through both the lines, so they are coplanar. $\endgroup$ Commented Jul 31, 2018 at 5:34
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    $\begingroup$ @PrinceKumar: it takes this into account: implicitly, all calculations really happen in the projective space $\mathbf P^3$, and the criterion checks whether there is or not a non-trivial solution in homogeneous coordinates. $\endgroup$
    – Bernard
    Commented Jul 31, 2018 at 8:02
  • $\begingroup$ what is the projective space you talked about? Can you please provide some geometric intuition why the determinant should be $0$? $\endgroup$ Commented Jul 31, 2018 at 12:10
  • $\begingroup$ Every affine space can be embedded in a projective space of the same dimension, adding a hyperplane at infinity. In such a space, affine parallel lines correspond to projective lines which intersect at $\infty$. You can see some details on Wikipedia's Projective space page. $\endgroup$
    – Bernard
    Commented Jul 31, 2018 at 12:20
  • $\begingroup$ @Bernand I am unable to understand the concept of Projective spaces. But here is one geometric way of thinking I arrived at: Directions 7,-4,7 and 4,3,-2 are perpendicular to the first line. So $span\{(7,-4,7), (4,3,-2)\}$ represents a plane through origin perpendicular to first line. Similarly, $span\{(1,-3,4), (1,-1,1)\}$ is perpendicular to the second line. Since the lines are parallel, both planes are same. So, $span\{(7,-4,7), (4,3,-2)\} = span\{(1,-3,4), (1,-1,1)\}$. Hence matrix formed from first 3 columns of the above determinant has rank 2. So, obviously the $rank< 4$. $\endgroup$ Commented Jul 31, 2018 at 13:01
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Disclaimer: This answer has become projective geometry in disguise.


By definition, for every point on the first line $L_1$ we have $$7x-4y+11z+16=0\qquad\text{ and }\qquad 4x+3y-2z+3=0.$$ It follows that for any nonzero linear combination of the coefficients of these planes, we get another plane that contains this line. That is to say, for all $\lambda_1,\mu_1\in\Bbb{R}$ with $(\lambda_1,\mu_1)\neq(0,0)$and all $(x,y,z)\in L_1$ we have $$(7\lambda_1+4\mu_1)x+(-4\lambda_1+3\mu_1)y+(11\lambda_1-2\mu_1)z+(16\lambda_1+3\mu_1)=0.$$ Conversely, every plane containing $L_1$ is given by such an equation for some $\lambda_1,\mu_1\in\Bbb{R}$ with $(\lambda_1,\mu_1)\neq(0,0)$. In this way, the set of planes in $\Bbb{R}^3$ containing $L_1$ corresponds to a plane in $\Bbb{R}^4$ with the origin removed.

Similarly, for the second line $L_2$, we see that for all $\lambda_2,\mu_2\in\Bbb{R}$ with $(\lambda_2,\mu_2)\neq(0,0)$ and all $(x,y,z)\in L_2$ we have $$(\lambda_2+\mu_2)x+(-3\lambda_2-\mu_2)y+(4\lambda_2-\mu_2)z+(6\lambda_2+\mu_2)=0.$$ And again conversely, every plane containing $L_2$ is given by such an equation for some $\lambda_2,\mu_2\in\Bbb{R}$ with $(\lambda_2,\mu_2)\neq(0,0)$, so the set of planes in $\Bbb{R}^3$ containing $L_2$ also corresponds to a plane in $\Bbb{R}^4$ with the origin removed.

Now the two lines are coplanar if and only if these two punctured planes in $\Bbb{R}^4$ intersect. This is equivalent to the two (unpunctured) planes intersecting in (at least) a line. This happens if and only if the determinant of $$\left[\begin{array}{rrrr} 7&-4&7&16\\ 4&3&-2&3\\ 1&-3&4&6\\ 1&-1&1&1 \end{array}\right]$$ is zero.

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    $\begingroup$ The lines can be parallel. They need not intersect. $\endgroup$ Commented Jul 31, 2018 at 5:31
  • $\begingroup$ You're absolutely right, I was thinking in the projective case. I'll improve my answer soon. $\endgroup$
    – Servaes
    Commented Jul 31, 2018 at 15:12
  • $\begingroup$ @amd You are right, I have corrected it. $\endgroup$
    – Servaes
    Commented Apr 12, 2019 at 12:03

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