Show that the cone whose vertex is at the
origin and which passes through the curve of
intersection of the sphere x^2 + y^2 + z^2 = 3p^2
and any plane which is at a distance 'p' from
the origin has mutually perpendicular
generators.
Answers
Answer:
Any plane at a distance 'p' from the origin is
lx+my+nz=p -------(1)lx+my+nz=p−−−−−−−(1)
(Hence l,m,nl,m,n are direction cosines of the normal to the plane). The equation of the cone whose vertex is the origin and which passes through the intersection of the given sphere of the plane (1) is;
x^2+y^2+z^2=3p^2=3(lx+my+nz)^2\\ \implies (1-3l^2)x^2+(1-3m^2)y^2+(1-3n^2)z^2-6lmxy-6mnyz-6nlxy=0----(2)x
2
+y
2
+z
2
=3p
2
=3(lx+my+nz)
2
⟹(1−3l
2
)x
2
+(1−3m
2
)y
2
+(1−3n
2
)z
2
−6lmxy−6mnyz−6nlxy=0−−−−(2)
With the equation of the cone generally defined as
f(x,y,z)=ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0f(x,y,z)=ax
2
+by
2
+cz
2
+2fyz+2gzx+2hxy=0
Then form
f(x,y,z)=(1-3l^2)x^2+(1-3m^2)y^2+(1-3n^2)z^2-6lmxy-6mnyz-6nlxy=0f(x,y,z)=(1−3l
2
)x
2
+(1−3m
2
)y
2
+(1−3n
2
)z
2
−6lmxy−6mnyz−6nlxy=0
and we get that
a=(1-3l^2);\\ b=(1-3m^2);\\ c=(1-3n^2)a=(1−3l
2
);
b=(1−3m
2
);
c=(1−3n
2
)
\therefore3-3(l^2+m^2+n^2)=3-3,1=0∴3−3(l
2
+m
2
+n
2
)=3−3,1=0
Hence the cone has three mutually perpendicular generators.