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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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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 $m_1n_2-m_2n_1$, $n_1l_2-n_2l_1$, $l_1m_2-l_2m_1$

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Toolbox:
  • If two lines are $\perp$ then the sum of the product of their direction cosines is 0.
  • $l_1l_2+m_1m_2+n_1n_2=0$
  • Sum of the squares of the direction cosines is 1.
Step 1:
It is given that $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are the direction cosines of two mutually perpendicular lines.
Therefore $l_1l_2+m_1m_2+n_1n_2=0$-------(1)
We also know that the sum of the squares of direction cosines is 1.
Therefore $l_1^2+m_1^2+n_1^2=1$-----(2)
$\qquad\;\;\;\;\; l_2^2+m_2^2+n_2^2=1$-----(3)
Let $l,m,n$ be the direction cosines of the line which is perpendicular to the line whose direction cosines are $l_1,m_1,n_1$ and $l_2,m_2,n_2$.
Therefore $ll_1+mm_1+nn_1=0$
$\qquad\;\;\;\;\;\;ll_2+mm_2+nn_2=0$
Step 2:
On solving the above equations,
$\large\frac{l}{m_1n_2-m_2n_1}=\frac{m}{n_1l_2-n_2l_1}=\frac{n}{l_1m_2-l_2m_1}$
On squaring we get
$\large\frac{l^2}{(m_1n_2-m_2n_1)^2}=\frac{m^2}{(n_1l_2-n_2l_1)^2}=\frac{n^2}{(l_1m_2-l_2m_1)^2}$
$\Rightarrow \large\frac{l^2+m^2+n^2}{(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2+(l_1m_2-l_2m_1)^2}$-----(4)
Step 3:
But $l^2+m^2+n^2=1$
We know that $(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)-(l_1l_2+m_1m_2+n_1n_2)^2=(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2+(l_1m_2-l_2m_1)^2$
From equ (1),equ(2) and eq(3) we get
$\large\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$
$\Rightarrow l=m_1n_2-m_2n_1$
$m=n_1l_2-n_2l_1$
$n=l_1m_2-l_2m_1$
Step 4:
Thus the direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$
answered Jun 3, 2013 by sreemathi.v
 

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