Answer to Question #94214 in Analytic Geometry for V Gaurav

Question

Find the angle between the lines whose direction cosines satisfy the equations l+m+n=0 and 2nl+2lm-mn=0.

Answer 1

#Task Q94214

$l+m+n=0$

$m=-(l+n)$

$2lm+2ln-mn=0$

$2ln+m(2l-n)=0$

$2ln-(l+n)(2l-n)=0$

$2ln-(2l^2-nl=2ln-n^2 )=0$

$-2l^2+nl+n^2=0$

$2l^2-nl-n^2=0$

$2l^2-2nl+nl-n^2=0$

$2l(l-n)+n(l-n)=0$

$(l-n)(2l+n)=0$

$l=n$

$l=-n/2$

$m=-(n+n)=-2n$

$n= -m/2$

$m=-(-n/2+n)$

$m=-n/2$

$l=-m/2=n/1$

$l/1=m/1=-n/2$

$l:m:n=1:-2:1$

$l:m:n=1:1:-2$

$cos \theta=\frac{1*1-2*1-2*1}{\sqrt{1^2+2^2+1^2}\sqrt{1^2+1^2+2^2}}=-\frac{1}{2}$

$θ=120^0$