Question

If l₁,m₁,n₁ and l₂,m₂,n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂-m₂n₁,n₁l₂-n₂l₁,l₁m₂-l₂m₁.

Answer 1

It is given that l₁,m₁,n₁ and l₂,m₂,n₂ are the direction cosines of two mutually perpendicular lines.

Therefore,

l₁l₂+m₁m₂+n₁n₂=0...(1)
l²₁+m²₁+n²₁=1...(2)
l²₂+m²₂+n²₂=1...(3)

Let l,m,n be the direction cosines of the line which is perpendicular to the line with direction cosines l₁,m₁,n₁ and l₂,m₂,n₂

∴ll₁+mm₁+nn₁=0 ll₂+mm₂+nn₂=0

∴\(\frac{l}{m_1n_2-m_2n_1}=\frac{m}{n_1l_2-n_2l_1}=\frac{n}{l_1m_2-l_2m_1}\)

\(\Rightarrow \frac{l^2}{(m_1n_2-m_2n_1)^2}=\frac{m^2}{(n_1l_2-n_2l_1)^2}=\frac{n^2}{(l_1m_2-l_2m_1)^2}\)

\(\Rightarrow \frac{l^2}{(m_1n_2-m_2n_1)^2}=\frac{m^2}{(n_1l_2-n_2l_1)^2}=\frac{n^2}{(l_1m_2-l_2m_1)^2}\)

=\(\Rightarrow \frac{l^2+m^2+n^2}{(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2+(l_1m_2-l_2m_1)^2}\).                                       ...(4)

l,m,n are the direction cosines of the line.
∴l²+m²+n²=1...(5)

It is known that,
(l²₁+m²₁+n²₁)(l²₂+m²₂+n²₂)-(l₁l₂+m₁m₂+n₁n₂)²=(m₁n₂-m₂n₁)²+(n₁l₂-n₂l₁)²+(l₁m₂-l₂m₁)²

From(1),(2),and(3),we obtain
\(\Rightarrow\) 1.1-0=(m1n2+m2n1)2+(n1l2+n2l1)+(l1m2+l2m1)
∴ (m₁n₂-m₂n₁)²+(n₁l₂-n₂l₁)²+(l₁m₂-l₂m₁)²=1....(6)

Substituting the values from equations(5)and(6) in equation(4), we obtain

\(\Rightarrow \frac{l^2}{(m_1n_2-m_2n_1)^2}=\frac{m^2}{(n_1l_2-n_2l_1)^2}=\frac{n^2}{(l_1m_2-l_2m_1)^2}=1\)

\(\Rightarrow\) l=m₁n₂-m₂n₁, m=n₁l₂-n₂l₁, n=l₁m₂-l₂m₁

Thus, the direction cosines of the required line are m₁n₂-m₂n₁, m=n₁l₂-n₂l₁, n=l₁m₂-l₂m_1.