Question

If the direction cosines of two lines are given by 3l +m+5n=0 and 6mn+2nl+5lm=0 ,
Find the angle between these two lines.

Answer 1

Member since Apr 11 2014

Sol: Given lines are 3l + m + 5n = 0  m = - (3l + 5n) -----------(1) and 6mn - 2nl + 5lm = 0 ----------(2) Substitute m in equation (2) we obtain ⇒ 6[- (3l + 5n)]n - 2nl + 5l[- (3l + 5n)] = 0 ⇒ -18ln - 30n2 - 2nl - 15l2 - 25ln = 0 ⇒ - 15l2 - 45ln - 30n2  = 0 ⇒ l2 + 3ln + 2n2  = 0 ⇒ l2 + 2ln + ln + 2n2  = 0 ⇒ l(l + 2n) + n(l + 2n)  = 0 ⇒ (l + n) (l + 2n) = 0 ∴ l = - n and l = -2n ( l / -1 ) = ( n / 1)  and ( l / -2) = ( n / 1) -------(3) Substitute l in equation we get m = - (3l + 5n) m = -2n  and m = n ( m / -2) = ( n / 1) and ( m / 1) = ( n / 1 )   --------(4) From ( 3) and (4) we get( l / -1 ) = ( m / -2) = ( n / 1) and ( l / -2) = ( m / 1) = ( n / 1 ) l : m : n = -1 : -2 : 1 and l : m : n = -2 : 1 : 1 i.e D.r's ( -1, -2, 1) and ( -2 , 1 , 1) Angle between the lines whose direction cosines are Cos θ = ( -1 × -2 + -2×1 + 1×1) / √ ((-1)2+ (-2)2 + 12)√ ((-2)2+ (1)2 + 12) Cos θ = 1 / √6 √6 Cos θ = 1 / 6 ∴ θ = cos-1 (1 / 6). ∴Angle between the lines whose direction cosines is cos-1 (1 / 6).