Question : Write an implicit equation of a plane which passes through the point with Cartesian coordinates (1, 2, 3) while being orthogonal to the straight line defined by $$ \left\{ \begin{split} x =& u + 1\\ y =& u – 2\\ z =& 2u + 1 \end{split} \right. $$ I had actually solved this months ago, but I do not recall how did I obtain the directional vector of [1 1 2], can anyone advise me on what did I do?
I know the steps after this, which are subbing into the equation of a(x-x1)+b(y-y1)+c(z-z1) which gives me x+y+2z-9 = 0 at the end, but I am unsure of how did I exactly get a b and c in this case.
The form of the line is $$ (x,y,z) = (u+1,u-2,2u+1) = u(1,1,2) + (1,-2,1),$$ where $(1,-2,1)$ is a point the line passes through, which determines which of many possible parallel lines the equations describe but makes no difference to whether a plane is orthogonal to the line. So we ignore $(1,-2,1)$.
On the other hand $u(1,1,2)$ tells us the line is parallel to the vector $(1,1,2).$
So you want something orthogonal to the vector $(1,1,2).$ One such plane is given by $(x,y,z) \cdot (1,1,2) = 0$ (where $\cdot$ is the inner product). And of course $(x,y,z) \cdot (1,1,2) = x+y+2z,$ so that's where your $a,b,c$ come from.
But your plane also has to pass through the given point. The rest of the math is finding the parallel plane that passes through that point.