Answer to Question #149883 in Analytic Geometry for Dhruv rawat

Question

There exists no line with 1/√3,1/√2, 1/√6 as direction cosines.
True or false with full explanation

Answer 1

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

Given statement is false.

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

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

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

Consider,

$L.H.S= l^2+m^2+n^2$

$= (\frac{1}{\sqrt{3}})^2+(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{6}})^2$

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

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

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

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

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

$= \frac{6}{6}$

$= 1$

$=L.H.S$

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

This is proved that, the given statement is false.