Question

Angle between two lines whose direction cosines are theorem

Answer 1

3l + m + 5n = 0

m = - (3l + 5n) -----------(1)

6mn - 2nl + 5lm = 0 ----------(2)

Substitute m=-(3l+5n) in eq(2)

⇒ 6[- (3l + 5n)]n - 2nl + 5l[- (3l + 5n)] = 0

⇒( -18ln - 30n)n-2nl-15l^2+25ln=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 1, 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),

( l / -2) = ( m / 1) = ( n / 1 )

l : m : n = -1 : -2 : 1

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+1^2))*√((-2)^2+1^2+1^2))

Cos θ = 1 / √6 √6

Cos θ = 1 / 6

∴ θ = cos inverse of (1/6)

∴Angle between the lines whose direction cosines is cos^-1(1/6)