Proving that two lines in three dimension given in asymmetric form (given as equations of two planes) are coplanar.

Question

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?

Answer 1

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 $0$.

Answer 2

Disclaimer: This answer has become projective geometry in disguise.

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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.