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Find the equation of the plane passing through $P(0,0,1)$ and containing $x=y=z$

a) y-z=0 b) x-z=0 c) z+x=1 d) x-y=0

My attempt:

I considered the point $P(0,0,1)$ for hypothesis and found two other points belonging to the line, for example, $(1,1,1)$ and $(0,0,0)$. So, solving the system:

$$\bigg \{ \begin{array}{rl} x+y+z = 0 \\ z = 0 \\ \end{array} $$

I get $(1,-1,0)$. Therefore, the answer could be d) x-y=0 or am I doing something wrong?

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    $\begingroup$ This looks fine! $\endgroup$ Jul 5, 2014 at 15:25

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What you do is correct. I just suggest you to use other letters for the coordinates of the plane that you are looking for. For example you can write your equation as $ax+by+cz=0$. You found two equations that you wrote $x+y+z=0$ and $z=0$ and correspond to replace the points $(1,1,1)$ and $(0,0,1)$ in the equation. These should be $a+b+c=0$ and $c=0$ and the solutions are $c=0$, $a=-b$, which gives you the plane given by $x-y=0$ (or any multiple of this equation).

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  • $\begingroup$ Nice hint, thanks! $\endgroup$ Jul 5, 2014 at 15:33
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    $\begingroup$ You are welcome. :-) $\endgroup$ Jul 5, 2014 at 15:53

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