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So, this question is more like two mini-questions that are subsets of a single regular-sized question. Say I have two planes: $x-z=1$ and $y+2z=3$. I'm trying to find their line of intersection.

a. Would it be okay to just take the lazy way out and add them, and end up with $2x+y=5$, which is the same as $y=-2x+5$? (I think probably not, because then $z$ might be unspecfied... or maybe $z$ is just $0$? Please tell me how wrong this whole train of reasoning is, and why, so that I can get a better understanding).

b. The explanation in my textbook told me to find a point on the line first and then take the cross product of the normal vectors of the planes, which would give me both a starting point and the line's direction. This sounds like a good idea, but the problem is that I'm not really sure how to find said point on the line. The book says to set $z$ equal to $0$ and then solve... but how would I know that the planes intersect at $z=0$ at all? What if the line I'm finding is parallel to the $xy$ axis or something? (And the same question for setting $x=0$, $y=0$, etc.)

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  • $\begingroup$ The line intersection represents all the points of two planes when the $z$ is same for these palnes( maybe the $z$ equal zero or maybe not) $\endgroup$ Commented Apr 16, 2015 at 8:39
  • $\begingroup$ @Essam But why/how? What if one plane's $z$ is another plane's $x$ (for example), or something else? $\endgroup$ Commented Apr 16, 2015 at 8:48

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For the first part of your question, adding the two planes does not yield their line of intersection. In fact, it does not even yield a line, it is the equation of a plane passing through their line of intersection.

In general, given two planes $P_1,P_2$, $$P_1+\lambda P_2=0, \lambda\in\mathbb{R}$$ represents the family of planes passing through the line of intersection of $P_1$ and $P_2$.

For the second part, taking the cross product would yield the direction ratios of the line. To find a point on the line easily generally one of $x,y$ or $z$ are set equal to zero, and the two equations are solved to get a point.

As you noticed, it is possible that the line does not intersect the plane $z=0$ in which case you can easily observe that the set of equations will be inconsistent.

In that case, put one of the other two as equal to zero. Obviously the line has to intersect two of the three planes $\{x=0,y=0,z=0\}$.

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  • $\begingroup$ how can you check that the line does not intersect the plane z = 0? $\endgroup$ Commented Jun 11, 2017 at 22:00
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Hint:

You can obtain the intersection line as: $$ \begin{cases} x-z=1\\ y+2z=3 \end{cases} $$ that has solution:

$$ \begin{cases} z=x-1\\ y=-2x+5 \end{cases} $$ so your line is: $$ (x,y,z)=(0,5,-1)+t(1,-2,1) $$ From this equation you can easely find a point on the line and his versor.

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