Let the direction cosines of two lines satisfy the equations 4l+m−n=0 and 2mn+10nl+3lm=0. Then the cosine of the acute angle between these lines is:
Solutions · 1
Official solution
From n=4l+m, substitute into 2mn+10nl+3lm=0:
\[2m(4l+m)+10l(4l+m)+3lm=0 \implies 40l^2+21lm+2m^2=0 \implies (8l+m)(5l+2m)=0.\]
Case 1: m=−8l, then n=4l−8l=−4l, direction ratios (1,−8,−4).
Case 2: 2m=−5l, i.e. l=2,m=−5 (scaled), then n=4(2)+(−5)=3, direction ratios (2,−5,3).
\[\cos\theta=\frac{(1)(2)+(-8)(-5)+(-4)(3)}{\sqrt{1+64+16}\cdot\sqrt{4+25+9}}=\frac{2+40-12}{9\sqrt{38}}=\frac{30}{9\sqrt{38}}=\frac{10}{3\sqrt{38}}.\]