Let the direction cosines of two lines satisfy the equations: \(4l+m-n=0\) and \(2mn+10nl+3lm=0\). Then the cosine of th…

Direction cosines of two lines satisfy the equation

\(4ℓ+m-n=0\\ \Rightarrow n=4ℓ+m...\left(1\right)\\ 2mn+10nℓ+3ℓm=0\\ \Rightarrow n\left(2m+10ℓ\right)+3ℓm=0...\left(2\right)\)

from eq. (1) and (2):

\(\Rightarrow (4ℓ+m)(2m+10ℓ)+3ℓm=0\\ \Rightarrow 8ℓm+40{ℓ}^{2}+2{m}^{2}+10ℓm+3ℓm=0\)

\(\Rightarrow 40{ℓ}^{2}+21ℓm+2{m}^{2}=0\\ \Rightarrow (8ℓ+m)(5ℓ+2m)=0\)

Case-1: \(8ℓ+m=0\Rightarrow m=-8ℓ\)
So direction ratio of \({L}_{1}\) is \(ℓ,-8ℓ,-4ℓ\)

Case-2: \(5ℓ+2m=0\Rightarrow m=\frac{-5}{2}ℓ\)
direction ratio of \({L}_{2}\) is \(ℓ,\frac{-5ℓ}{2},\frac{3ℓ}{2}\)

\(\cos \theta =\left|\frac{{ℓ}^{2}+20{ℓ}^{2}-6{ℓ}^{2}}{\sqrt{{ℓ}^{2}+64{ℓ}^{2}+16{ℓ}^{2}}\sqrt{{ℓ}^{2}+\frac{25{ℓ}^{2}}{4}+\frac{9{ℓ}^{2}}{4}}}\right|\\ =\frac{15{ℓ}^{2}}{\left(9ℓ\right)\frac{\sqrt{38}ℓ}{2}}=\frac{10}{3\sqrt{38}}\\ =\frac{10}{3\sqrt{38}}\)