Question

Angle between two lines whose direction cosines satisfy the equation 3l+m+5n=0, 6mn-2nl+5lm=0

Answer 1

The two lines are perpendicular to each other. Explanation: Let the direction cosines of the two lines be (l1,m1 ,n1) and (l2 .... 2n(l+n)+nl+2l(l+n)=0 or 2l2+5ln 

Answer 2

The two equations are,

3 l+m+5 n=0,⇒ m= -( 3 l + 5 n)

Putting the value of m in the second equation, which is

→6 m n-2 n l+5 l m=0

→ 6 m n + 5 l m - 2 n l = 0

→ m ( 6 n + 5 l) - 2 n l =0

→ -(3 l + 5 n)( 6 n + 5 l) - 2 n l =0

Dividing by (-) sign

→18 n l + 15 l² + 30 n² + 25 n l + 2 n l =0

→ 15 l²+ 45 n l+ 30 n²= 0 [Adding like terms]

Dividing by 15 on both sides

→ l² + 3 n l  + 2 n²=0

Splitting the middle term

→ l² + 2 n l + n l + 2 n²=0

→ l ( l+ 2 n) + n (l + 2 n)=0

→ (l+n)(l+2 n)=0

→ l=-n, l=-2 n

Also , m=-(3 l + 5 n)

Substituting the values of l, we get ,when

l= -n,→ m =-(3 ×-n + 5 n)→m= - 2 n,

And when , l=- 2 n, m= -(3× -2 n + 5 n)= -( -6 n+ 5 n)= n

l= -n , m= - 2 n → l/-1  =m/-2=n/1  .................(1)

l= -2 n, m=n→ l/-2 = m/1  = n/1  .........................(2)

As direction ratios are proportional to Direction cosines

So, (-1, -2,1) and (-2,1,1) are direction ratios of the line.

Angle between the two lines whose direction cosines or ratios are (a,b,c) and (p,q,r) is given by Cosα.The formula is

Cosα= $\frac{ap +bq +cr}{ \sqrt {a^{2}+b^{2}+c^{2}}\times \sqrt {p^{2}+q^{2}+r^{2}}}$

Cos α = $\frac{[ (-1)\times(-2)+(-2)\times(1)+ (1)\times(1)]}{ {\sqrt {1+4+4}}\times {\sqrt{4+1+1}}}$

Cos α = $\frac{1}{\sqrt6^{2}}=\frac{1}{6}$

Angle between two lines  whose direction cosines satisfy the equation 3l+m+5n=0, 6mn-2nl+5lm=0 is $Cos^{-1}\frac{1}{6}$