Answer to Question #168339 in Analytic Geometry for Anshika

Question

The plane lx+my+nz=0 moves in such a way that it's intersection with the planes ax+by+cz+d=0 , a'x+b'y+c'z+d'=0 are perpendicular. Show that the normal to the plane through the origin describes, in general, a cone of second degree , and find its equation . Examine the case when aa'+bb'+cc'=0 , that is , when the two given planes are perpendicular.

Answer 1

The plane $lx+my+nz=0$ moves in such a way that it's intersection with the planes $ax+by+cz+d=0$ , $a'x+b'y+c'z+d'=0$ are perpendicular

The intersection with first plane gives the following line,

$\dfrac{x}{cm - bn} = \dfrac{y}{an - cl} = \dfrac{z}{bl - am}$

Similarly the intersection with the second line is :

$\dfrac{x}{c'm - b'n} = \dfrac{y}{a'n - c'l} = \dfrac{z}{b'l - a'm}$

For these lines to be perpendicular the direction ratio's inner product should be 0 , i.e.

$(cm - bn)(c'm - b'n) + (an - cl)(a'n - c'l) + (bl - am)(b'l - a'm) = 0$

Hence, this is equation of cone of second degree.