An angle between the lines whose direction cosines are given by the equations l+3m+5n=0 and 5lm−2mn+6nl=0, is
Solutions · 1
Official solution
From l=−3m−5n, substituting into the second equation:
\[5(-3m-5n)m-2mn+6n(-3m-5n)=0\implies m^2+3mn+2n^2=0\implies (m+n)(m+2n)=0.\]
This gives (l,m,n)=(−2,−1,1) or (1,−2,1) (up to scale).
\[\cos\theta=\frac{(-2)(1)+(-1)(-2)+(1)(1)}{\sqrt{6}\cdot\sqrt{6}}=\frac16.\]