Questions:
1) In analytical geometry, normal vector to a plane is found by the cross product of two lines in that plane, i wonder what the magnitude of the normal vector to the plane tells us (how it's determined)?
2) How the dot product of a vector from origin pointing a point in the plane and normal to that plane gives us the distance of the plane from the origin ?
r.n = d
where 'r' is the position vector of a point lying on the said plane, 'n' is the unit normal vector parallel to the normal that joins the origin to the plane, 'd' is the perpendicular distance of the plane from the origin.
I can't understand the relationship of 'd' here, please explain.
Thanks in advance!
The equation of a 3D plane is $$a x + b y + c z - d = 0$$ In vector notation, using $\vec{n} = (a, b, c)$ and $\vec{r} = (x, y, z)$, $$\vec{n} \cdot \vec{r} - d = 0$$ which we can read as "Point $\vec{r}$ is on the plane if and only if the length of its projection to the plane normal $\vec{n}$ is $d$."
Because the shortest distance is measured along the normal, $d$ is the signed distance from origin to the plane, in units of the length of the normal vector.
If $\vec{n}$ is an unit vector, then $d$ is the signed distance. If the plane is in the direction of the normal vector, then $d \gt 0$; if the plane is in the opposite direction, then $d \lt 0$. If the plane passes through origin, $d = 0$.
So, the answers to the stated questions are:
The length of the normal vector determines the scale for the signed distance from origin, $d$. If say $\lVert\vec{n}\rVert = 2$, then $d$ is half the measured distance.
Shortest distance is always measured along the surface normal. The minimum distance between plane and the origin is measured along the plane normal, along the line passing through origin.