Question #149883
There exists no line with 1/√3,1/√2, 1/√6 as direction cosines.
True or false with full explanation
1
Expert's answer
2020-12-15T19:43:08-0500

Solution: Given that "There exists no line with 13,12,16\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{6}} as direction cosines."

Given statement is false.

Since we have "The direction cosines are related with l2+m2+n2=1,l^2+m^2+n^2=1,

where l,m,nl,m,n are direction cosines. "

Here consider l=13,m=12,n=16l=\frac{1}{\sqrt{3}},m=\frac{1}{\sqrt{2}} ,n=\frac{1}{\sqrt{6}}

Consider,

L.H.S=l2+m2+n2L.H.S= l^2+m^2+n^2


=(13)2+(12)2+(16)2= (\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{6}})^2


=(13)+(12)+(16)= (\frac{1}{3})+(\frac{1}{2})+(\frac{1}{6})

=(13+12)+16= (\frac{1}{3}+\frac{1}{2})+\frac{1}{6}

=(2+3(3)(2))+16= (\frac{2+3}{(3)(2)})+\frac{1}{6}


=56+16= \frac{5}{6}+\frac{1}{6}

=5+16= \frac{5+1}{6}

=66= \frac{6}{6}

=1= 1


=L.H.S=L.H.S


Therefore , there exists a line with 13,12,16\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{6}} as direction cosines.

This is proved that, the given statement is false.


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