Solution: Given that "There exists no line with 1 3 , 1 2 , 1 6 \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{6}} 3 1 , 2 1 , 6 1 as direction cosines."
Given statement is false .
Since we have "The direction cosines are related with l 2 + m 2 + n 2 = 1 , l^2+m^2+n^2=1, l 2 + m 2 + n 2 = 1 ,
where l , m , n l,m,n l , m , n are direction cosines. "
Here consider l = 1 3 , m = 1 2 , n = 1 6 l=\frac{1}{\sqrt{3}},m=\frac{1}{\sqrt{2}} ,n=\frac{1}{\sqrt{6}} l = 3 1 , m = 2 1 , n = 6 1
Consider,
L . H . S = l 2 + m 2 + n 2 L.H.S= l^2+m^2+n^2 L . H . S = l 2 + m 2 + n 2
= ( 1 3 ) 2 + ( 1 2 ) 2 + ( 1 6 ) 2 = (\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{6}})^2 = ( 3 1 ) 2 + ( 2 1 ) 2 + ( 6 1 ) 2
= ( 1 3 ) + ( 1 2 ) + ( 1 6 ) = (\frac{1}{3})+(\frac{1}{2})+(\frac{1}{6}) = ( 3 1 ) + ( 2 1 ) + ( 6 1 )
= ( 1 3 + 1 2 ) + 1 6 = (\frac{1}{3}+\frac{1}{2})+\frac{1}{6} = ( 3 1 + 2 1 ) + 6 1
= ( 2 + 3 ( 3 ) ( 2 ) ) + 1 6 = (\frac{2+3}{(3)(2)})+\frac{1}{6} = ( ( 3 ) ( 2 ) 2 + 3 ) + 6 1
= 5 6 + 1 6 = \frac{5}{6}+\frac{1}{6} = 6 5 + 6 1
= 5 + 1 6 = \frac{5+1}{6} = 6 5 + 1
= 6 6 = \frac{6}{6} = 6 6
= 1 = 1 = 1
= L . H . S =L.H.S = L . H . S
Therefore , there exists a line with 1 3 , 1 2 , 1 6 \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{6}} 3 1 , 2 1 , 6 1 as direction cosines.
This is proved that, the given statement is false.
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