The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).

I know that for two mutually perpendicular lines $$l_1l_2+m_1m_2+n_1n_2=0.$$ But I don't know the further what to do Please can anyone guide me further? Thank you.

  • $\begingroup$ To do a complete proof you have to show that those quantities are not all zero -- one or two may be zero -- using the fact that l1, m1, n1 are not all zero, l2, m2, n2 are not all zero, and that the lines are perpendicular. Once you have done that, use the same test you already know to compare that third line to each of the other two. $\endgroup$ – EGoodman Aug 3 '18 at 4:40
  • $\begingroup$ You may find this helpful, as they're essentially attempting to get you to reinvent the cross product: en.wikipedia.org/wiki/Cross_product#Geometric_meaning $\endgroup$ – EGoodman Aug 3 '18 at 4:40
  • $\begingroup$ Thank you very much sir I got the answer by your way . $\endgroup$ – Ritik Aug 3 '18 at 4:46

Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.

$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.

Since $u_1\times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1\times u_2$ $$u_1\times u_2=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ $$=\vec{i}(m_1n_2-m_2n_1)-\vec{j}(n_2l_1-n_1l_2)+\vec{k}(l_1m_2-l_2m_1)$$ $$=\vec{i}(m_1n_2-m_2n_1)+\vec{j}(n_1l_2-n_2l_1)+\vec{k}(l_1m_2-l_2m_1)$$ Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$

  • $\begingroup$ Thanks I got the answer $\endgroup$ – Ritik Aug 3 '18 at 4:48

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