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-l2m1.