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Question

# Find the angle between the lines whose direction cosines satisfy the equations ℓ+m+n=0, ℓ2+m2−n2=0

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Solution

## Given l+m+n=0........................(1) and l2+m2−n2=0.....................(2) By equation (1) l+m+n=0 l+m=−n −(l+m)=n n=−(l+m) put in equation (2) l2+m2−n2=0 l2+m2−(−(l+m))2=0 l+m−(l2+m2+2lm)=0 l2+m2−l2−m2−2lm=0 then −2lm=0 lm=0 Either l=0 or m=0 Let us put l=0 in equation (1), and we get l+m+n=0 0+m+n=0 m=−n Direction ratios (l,m,n)=(0,1,−1) Let us put m=0 in equation (1) and we get l+m+n=0 l+0+n=0 l=−n Direction ratios (l,m,n)=(1,0,−1) We know that angle between of two given line cosθ=b1.b2|b1||b2| First solve b1.b2 Then b1.b2=(1,0,−1)(0,1,−1) b1.b2=0+0+1 b1.b2=1 And |b1|=√02+12+(−1)2=√2 |b2|=√02+12+(−1)2=√2 Now substituting the value in cosθ=b1.b2|b1||b2| cosθ=1√2√2 cosθ=12 cosθ=cosπ3 θ=π3 Hence it is required solution

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