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Question

# Find the angle between the two straight lines whose direction cosines are given by 2l+2m−n=0 and mn+nl+lm=0.

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Solution

## Let the angle between the lines be αGiven Direction cosines of two lines are <l,m,n> Also given that 2l+2m−n=0⟹n=2l+2m⋯(1)mn+nl+lm=0⋯(2)⟹m(2l+2m)+l(2l+2m)+lm=0 (∵from(1) )⟹2m2+5lm+2l2=0⟹2m2+4lm+lm+2l2=0⟹2m(m+2l)+l(m+2l)=0⟹(2m+l)(m+2l)=0⟹m=−l2,−2ln=2l+2m=2l+2(−l2),2l+2(−2l)=2l−l,2l−4l=l,−2lSo the direction cosines of two lines are <l,−l2,l> and <l,−2l,−2l>So the direction ratios will be <2,−1,2> and <1,−2,−2>As we know thatIf θ is the angle between the two lines having direction ratios as <a,b,c> and <d,e,f> then cosθ=ad+be+cf√a2+b2+c2√d2+e2+f2Hence cosα=(2)(1)+(−1)(−2)+2(−2)√22+(−1)2+22√12+(−2)2+(−2)2=2+2−43×3=0⟹α=π2So the angle between the two lines is π2

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