Question

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Find the angle between the lines whose direction cosines are given by the equation
3l + m + 5n = 0 , 6mn - 2nl +5lm = 0

Answer 1

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 : 1i.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).

Answer 2

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Class-XII Maths

Murtaza Nimachwala

Dec 10, 2014

Find the angle between the lines whose direction cosines are given by the equations ?

Find the angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0 and 6mn - 2nl + 5lm = 0

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Raghunath Reddy

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).