Question
Let $$L$$ be the line of intersection of the planes $$2x+3y+z=1$$ and $$x+3y+2z=2.$$ If $$L$$ makes an angle $$\alpha $$ with the positive $$x$$-axis, then $$\cos \,\alpha $$ equals :
A.
$$1$$
B.
$$\frac{1}{{\sqrt 2 }}$$
C.
$$\frac{1}{{\sqrt 3 }}$$
D.
$$\frac{1}{2}$$
Answer :
$$\frac{1}{{\sqrt 3 }}$$
Solution :
Let the direction cosines of line $$L$$ be $$l,\,m,\,n$$ then
$$2l + 3m + n = 0.....({\text{i}})$$
and $$l + 3m + 2n = 0.....({\text{ii}})$$
on solving equation (i) and (ii), we get
$$\frac{l}{{6 - 3}} = \frac{m}{{1 - 4}} = \frac{n}{{6 - 3}}\,\,\, \Rightarrow \frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3}$$
$$\eqalign{
& {\text{Now }}\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }} \cr
& \because \,{l^2} + {m^2} + {n^2} = 1 \cr
& \therefore \,\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{1}{{\sqrt {27} }} \cr
& \Rightarrow l = \frac{3}{{\sqrt {27} }} = \frac{1}{{\sqrt 3 }},\,\,m = - \frac{1}{{\sqrt 3 }},\,\,n = \frac{1}{{\sqrt 3 }} \cr} $$
Line $$L,$$ makes an angle $$\alpha $$ with $$+ve\, x$$ -axis
$$\therefore l = \cos \,\alpha \,\,\,\, \Rightarrow \cos \,\alpha = \frac{1}{{\sqrt 3 }}$$