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 $i=1,2,3$.

From the above relation, we have
$l_1m_1+l_2m_2+l_3m_3=0$ 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?

  • $\begingroup$ Your approach is correct but overcomplicated: see the answer below. $\endgroup$ – Aretino May 26 at 14:12

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.


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