- #36
Physicsissuef
- 908
- 0
Hootenanny said:Note that usually r is the position vector of a general point such that r = xi+yj+zk. Hence, if n = ai+bj+ck then,
[tex]\underline{r}\cdot\underline{n} - p = 0[/tex]
[tex]\left(x\underline{i} + y\underline{j} + z\underline{k}\right)\cdot\left(a\underline{i} + b\underline{j} + c\underline{k}\right)-p=0[/tex]
[tex]ax+by+cz - p =0[/tex]
Which is an equation of a plane in three space.
But again, something is wrong, since if p=|n|. In this case it isnt.