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Question:

An angle between the lines whose direction cosines are given by the equations, l + 3m + 5n = 0 and 5lm - 6mn + 6nl = 0, is

cos⁻¹(18)

cos⁻¹(16)

cos⁻¹(14)

cos⁻¹(13)

Solution:

Givenl+3m+5n=0..1and5lm𕒶mn+6nl=0...2Herel,m,nare directional cosines.From 1,l=𕒷m𕒹nSubstituting equation 1 in equation 25(𕒷m𕒹n)m𕒶mn+6n(𕒷m𕒹n)=015m2+45mn+30n2=0⇒m2+3mn+2n2=0⇒m2+2mn+mn+2n2=0⇒(m+n)(m+2n)=0∴m=−norm=𕒶nForm=−n;l=𕒶nAnd form=𕒶n;l=n∴(l,m,n)=(𕒶n,−n,n)Or(l,m,n)=(n,𕒶n,n)⇒(l,m,n)=(𕒶,𕒵,1)or⇒(l,m,n)=(1,𕒶,1)cos(θ)=A.B|A||B|,θis angle between the lines⇒cos(θ)=𕒶.1+(𕒵) (𕒶)+1.1√6.√6⇒cos(θ)=16∴θ=cos𕒵(16)Hence, correct option is 'B'