(r-rP) . n = 0
has normal vector n=(1,1,2) and goes through point rP.
So this plane and x + y + 2z = 1 have the same normal vectors are parallel.
I've been looking through past exam papers for an exam tomorrow and 2 questions I've encountered have been along the lines of this one:
Write out the Cartesian equation of a plane passing through the point
with the position vector rP = (1, 1, 1) and perpendicular to the plane
whose Cartesian equation is x + y + 2z = 1.
If there is more than one such plane, you can present equation of any
one of them.
I tried doing this by using the normal as (1, 1, 2) and dotting it with (x-1, y-1, z-1) and putting it equal to 0 (using (r-rP) . n = 0 ) but I dont think my answer was right.
(The paper obviously didn't put all answers as there are infinitely many such planes but one example answer was 2x-z=1)
(Also just as an addition one of the other questions practically the same to this one said to just write the equation of the plane (not Cartesian) and the answer was in the form n = (answer) - I thought that was the vector normal not an equation of the plane?)
Any help on this is really appreciated! Thanks