Answer to Question #94214 in Analytic Geometry for V Gaurav

Question #94214
Find the angle between the lines whose direction cosines satisfy the equations l+m+n=0 and 2nl+2lm-mn=0.
1
Expert's answer
2019-09-12T05:10:32-0400

#Task Q94214

l+m+n=0l+m+n=0

m=(l+n)m=-(l+n)

2lm+2lnmn=02lm+2ln-mn=0

2ln+m(2ln)=02ln+m(2l-n)=0

2ln(l+n)(2ln)=02ln-(l+n)(2l-n)=0

2ln(2l2nl=2lnn2)=02ln-(2l^2-nl=2ln-n^2 )=0

2l2+nl+n2=0-2l^2+nl+n^2=0

2l2nln2=02l^2-nl-n^2=0

2l22nl+nln2=02l^2-2nl+nl-n^2=0

2l(ln)+n(ln)=02l(l-n)+n(l-n)=0

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

l=nl=n

l=n/2l=-n/2

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

n=m/2n= -m/2

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

m=n/2m=-n/2

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

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

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

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

cosθ=11212112+22+1212+12+22=12cos \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=1200θ=120^0


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