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Let us consider two planes with equations:

$P_1 : x + 2y + 3z = 3$ and $P_2 : 2x-y = 5$

By substituting $x$ as $0$, can I say that the point $(0,-5,13/3)$ lies on the line of intersection of two planes?

And if I find that the direction vector of the line of intersection is $(3,5,-5)$, Can I say that the equation of the line of intersection of two planes is $(0,-5,13/3) + t(3,5,-5)$ ?

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  • $\begingroup$ You can partially check your own work: The direction vector of the line must be perpendicular to the normals of both planes (why?). Is it? $\endgroup$ Commented Feb 11, 2019 at 3:21
  • $\begingroup$ Yes, the direction vector was calculated by the cross product of the normals. So it must be perpendicular to the line. $\endgroup$ Commented Feb 11, 2019 at 15:14
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    $\begingroup$ It should be perpendicular, but it is really? One can see at a glance that $2\cdot3-5\ne 0$. Asserting that the vector must be perpendicular because it’s the result of a cross product isn’t at all the same as actually checking that it is: you must use a different method than the one you used to generate the value in the first place, since otherwise you’re likely to make the same mistakes (if any). $\endgroup$ Commented Feb 11, 2019 at 18:26
  • $\begingroup$ "Can I say that the equation of the line of intersection of two planes is $(0,-5,13/3) + t(3,5,-5)$ ?" Even if you fixed the incorrect direction vector, the answer would still be no, because that's not an equation. What you do have is a parametric form of a line. $\endgroup$ Commented Nov 4, 2021 at 12:42

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I think in the vector $(3,5,-5)$ there is a mistake.

I like the following way.

Let $x=t$.

Thus, $$y=-5+2t,$$ $$z=\frac{13}{3}-\frac{5}{3}t,$$ which gives the answer $$\left(0,-5,\frac{13}{3}\right)+t(3,6,-5).$$

I got it by the following reasoning: $$x=0+1\cdot t,$$ $$y=-5+2t$$ and $$z=\frac{13}{3}-\frac{5}{3}t$$ or since $t$ is any real number, $$x=0+3t,$$ $$y=-5+6t$$ and $$z=\frac{13}{3}-5t$$

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  • $\begingroup$ What did you do after representing x,y and z in terms of t? Yea I think the 5 should be replaced with 6, my mistake $\endgroup$ Commented Feb 10, 2019 at 8:06
  • $\begingroup$ And is the point on the line correct? Is it ok to take x as 0 like i did? $\endgroup$ Commented Feb 10, 2019 at 8:06
  • $\begingroup$ The point is correct. $\endgroup$ Commented Feb 10, 2019 at 8:13

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