1
$\begingroup$

Suppose, we have two intersecting planes $P: ax+by+cz+d=0$ and $P':a'x+b'y+c'z+d'=0$ where $d, d'$ have same sign. Then the origin lies either in acute angle side(i.e. where the angle between the planes is acute) or in the obtuse angle side(i.e. where the angle between the planes is obtuse).Two plane intersecting

I can prove that the origin lies on the acute side iff $aa'+bb'+cc'<0$.
It can be shown that the two planes bisecting the angle are-$$\frac{ax+by+cz+d}{\sqrt{a^2+b^2+c^2}}=\pm\frac{a'x+b'y+c'z}{\sqrt{a'^2+b'^2+c'^2}}$$ Now I will calculate the angle $\theta$ between one of the bisecting planes and one given plane(say $P$). If $\theta<45^\circ$, then that bisecting plane will be on the acute angle side, otherwise will be on the obtuse angle side.
So, to find the which of these two bisecting planes lie on the same side where the origin lies, I will follow the steps:
(1) I will find where the origin lies by checking the sign of $aa'+bb'+cc'$.
(2) I will check which one of the bisecting planes lie on the acute angle side and which one lies on the obtuse angle side by calculating the angle $\theta$.
(3) Now, I can easily find out the required plane by the results of step (1) and (2). If $aa'+bb'+cc'<0$ then the answer will be the plane which is on the acute side otherwise the answer will be the plane which is on the obtuse side.
But I found in many cases it is difficult to calculate $\theta$ by hand. So, there is a small hint in my book which says-

If $aa'+bb'+cc'<0$, the plane $\frac{ax+by+cz+d}{\sqrt{a^2+b^2+c^2}}=+\frac{a'x+b'y+c'z}{\sqrt{a'^2+b'^2+c'^2}}$ will be the required plane (i.e. it lies on the acute side), and if $aa'+bb'+cc'>0$, the plane $\frac{ax+by+cz+d}{\sqrt{a^2+b^2+c^2}}=\frac{a'x+b'y+c'z}{\sqrt{a'^2+b'^2+c'^2}}$ will be the required plane (i.e. it lies on the obtuse side).


But I can't prove this hint. Can anybody prove this hint? Thank for your patience and help in advance.

$\endgroup$
0
$\begingroup$

The normal vectors $(a,b,c)$ and $(a',b',c')$ point in the direction of increase of the corresponding linear function $F(x)=ax+by+cz+d$, $G(x)=a'x+b'y+c'z+d'$. Since $d$ and $d'$ have the same sign, at the origin both functions are either positive or negative. Suppose they are both positive. Then, the origin will be on the side of the planes towards which $(a,b,c)$ (respectively $(a',b',c')$) points. You can easily convince yourself that when $aa'+bb'+cc'<0$ (that is, when the angle between the normal vectors is obtuse) the values of both $F$ and $G$ increase as you move into the acute angle, hence your bisector is the one corresponding to Geometric level of $F$ = Geometric level of $G$ (this is what the bisector equation means: you set "normalized" levels to be equal). When $aa'+bb'+cc'>0$ (that is, when the angle between the normal vectors is acute) the values of $F$ and $G$ change in opposite directions as you move into the acute angle, hence the origin is not there: it is in the obtuse sector, and the correspoding bisector is Normalized level of $F$= - Normalized level of $G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.