I'm working on this question
Find the perpendicular distance from the origin to the plane $x+2y+2z=6$. I looked into the solution and it says the plane has a normal vector $(1,2,2)$ due to the coefficient of the plane equation. Then it picked an arbitrary point $P =(6,0,0)$ on the plane and use the projection equation of PO onto the vector $<1,2,2>$ to solve for distance.
I understand the logic for using the projection equation, but my confusion is that, how do we know the normal vector from the plane starts at the point $(6,0,0)$, if we want to project $PO$ on the vector, doesn't vector need to start at point $P$?
Finding length of a vector that is normal to a plane to the origin
linear algebramultivariable-calculus
Best Answer
I think that is simple; let's talk about the general case. Choose a point $P(x,y,z)$ on the plane $\Pi:ax+by+cz=1$, which has a normal vector $\vec{n_1}=(r,s,t)$. ($r,s,t$ can be easily calculated.)
Since $\vec{OP}\parallel\vec{n_1}$, we let $x=ru,y=su,z=tu$. And, because $P$ is on $\Pi$, we have $ax+by+cz=1$, or $aru+bsu+ctu=1$. So $u=\frac{1}{ar+bs+ct}$, $P(\frac{r}{ar+bs+ct},\frac{s}{ar+bs+ct},\frac{t}{ar+bs+ct})$. Length can be calculated easily then.