# If a right circular cone has three mutually perpendicular generators then find its semi-vertical angle.

If a right circular cone has three mutually perpendicular generators then find its semi-vertical angle.

We see that if $$ax^2+by^2+cz^2+2fyx+2gzx+2hxy=0$$ has three mutually perpendicular generators, then $$a+b+c=0$$. But I don't know what will be the way to find the semi-vertical angle.

Let us consider three mutually perpendicular generators with direction cosines $$l_i,m_i,n_i$$ for $$i=1,2,3$$. The direction cosines of the axis are $$\frac{\sum l_1}{3},\frac{\sum m_1}{3},\frac{\sum n_1}{3}=l',m',n'$$ (say) Since these three generators are mutually perpendicular, we have
$$l_il_j+m_im_j+n_in_j=0$$ for $$i\neq j$$. Also we can say that $$l_i^2+m_i^2+n_i^2=1$$ for $$i=1,2,3$$.
$$l_1m_1+l_2m_2+l_3m_3=0$$ etc. $$\cos\alpha=\frac{l_1l'+m_1m'+n_1n'}{\sqrt{l'^2+m'^2+n'^2}}=\frac{1}{\sqrt{3}}\implies \alpha=\tan^{-1}(\sqrt{2})$$. Is my approach correct?
If the generators are in the directions of the usual coordinate axes, then the axis of the cone is in the direction of the vector $$(1,1,1)$$. So the semi-vertical angle is the angle between the vectors $$(1,0,0)$$ and $$(1,1,1)$$ etc.