We have been studying the Hesse normal form of the plane equation, but the sketch of the plane in space given by the lecturer was horrible.
Basically I ask you to explain me how does one obtain the Hesse normal form equation geometrically.
Also I need clearing up on the question of,why is the unit vector n0 of vector normal to the plane equal to cosines of the angles with axes.
Once again I do not need the process of conversion from general cartesian form into Hesse form, I need geometric as well as intuitive explanation.
Hope I was clear enough because the resources on the subject on the Web are quite scarce.
Thank you in advance.
Best Answer
Ok here goes,self answer QA style baby.
First introduce some notation :
Let P be our plane
Let $\bar n$ be vector orthogonal to plane and let $d=|\bar n|$
Let $M\in P$ be arbitrary point M(x,y,z),let $\bar r$ be position vector of point M
Let $pr_x y = \frac{x \dot y}{|y|}$ be projection of y onto x
Let $\alpha,\beta,\gamma$ be angles that $\bar n $ makes with x,y,z axes respectively
Now we can consider the unit vector $\bar n_0$ and we can conclude that
$cos(i,\bar n_0 ) = cos \alpha = \frac{i \bar n_0}{|i||\bar n_0|}= \frac{pr_i \bar n_0}{|\bar n_0|} = pr_i \bar n_0$
Same for beta and delta,thus we can conclude that $\bar n_0 = \langle cos \alpha , cos\beta,cos \gamma \rangle$
Now we can consider the point L at distance d from the origin(connected with origin by orthogonal vector)
Then triangle OLM is right triangle and thus $pr_{ \bar n} \bar r$ is equal to d and from this conclusion we obtain the form in following steps:
$$pr_{\bar n} \bar r = \frac{\bar r \bar n }{|n|} = \frac{d \bar r \bar n_0}{d|\bar n_0|}=d$$
$$\bar r \bar n_0 = d$$
And that is the way its done