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I am trying to answer the question "find a plane that passes through point (0,0,0) and is perpendicular to the plane x+2y+3z=6."

I am suspecting that there are infinitely many such planes. Is that correct?

Here is how I am thinking about it. A vector that is normal to the given plane is [1,2,3]. For the target plane, if [a,b,c] is a vector that is normal to it, we must have the dot product of vectors [1,2,3] and [a,b,c] equal zero, that is, a+2b+3c = 0. This equation is satisfied for many combinations of values of a, b, and c.

Two such possibilities are: a=1, b=1, and c=(-1), OR a=1, b=(-2), and c=1. Considering an arbitrary point (x,y,z) on the target plane and taking into account that the target plane passes through (0,0,0), the vector [x-0, y-0, z-0] or [x,y,z] is parallel to the target plane. Since [a,b,c] is normal to the target plane, dot product of [a,b,c] and [x,y,z] must be zero, that is, ax + by + cz = 0.

So, substituting the two different combinations of values for a, b, and c, we arrive at two different equations identifying two different planes that satisfy the requirements: either x + y - z = 0 OR x - 2y + z = 0.

I am not sure if I am correct to say that there are infinitely many distinct planes that are solutions to this question.

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    $\begingroup$ Of course. There is a line that is perpendicular to the plane and passes through the origin. Any plane containing that line works $\endgroup$ – Will Jagy Jun 1 '16 at 2:54
  • $\begingroup$ This exact same question was asked earlier today but I do not know how to locate it. $\endgroup$ – John Wayland Bales Jun 1 '16 at 2:57
  • $\begingroup$ Your reasoning is correct, there are infinitely many correct answers. If $a+2b+3c=0$ then the plane $ax+by+cz=0$ is perpendicular to the plane $x+2y+3z=6$ and contains $(0,0,0)$. $\endgroup$ – John Wayland Bales Jun 1 '16 at 3:02

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