Question

What is the angle between two lines whose direction ratios satisfy following equations?

Find the angle between two lines whose direction ratios satisfy `2l-m+2n=0` and `mn+nl+lm=0`

Answer 1

The two lines are perpendicular to each other.

Explanation:

Let the direction cosines of the two lines be `(l_1,m_1,n_1)` and `(l_2,m_2,n_2)` and as they satisfy the conditions we have

`2l-m+2n=0` .............................(1) and

`mn+nl+lm=0` .............................(2)

Note that if `(a_1,b_1,c_1)` and `(a_2,b_2,c_2)` are direction ratios of two lines, we have `l_1/a_1=m_1/b_1=n_1/c_1` and `l_2/a_2=m_2/b_2=n_2/c_2`

and the angle `theta` between them is given by

`costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))`

From (1), we get `m=2(l+n)` and substituting in (2) we get

`2n(l+n)+nl+2l(l+n)=0` or `2l^2+5ln+2n^2=0`

or `(l+2n)(2l+n)=0`

i.e either `l+2n=0` or `2l+n=0`

if `l+2n=0`, `l=-2n` and `m=2(-2n+n)=-2n` and we have

`l_1/(-2)=m_1/(-2)=n_1/1`

and if `2l+n=0`m `n=-2l` and `m=2(l-2l)=-2l` and we have

`l_2/(1)=m_2/(-2)=n_2/(-2)`

and `costheta=(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))`

= `((-2)xx1+(-2)xx(-2)+1xx(-2))/(sqrt(4+4+1)sqrt(1+4+4))`

= `(-2+4-2)/9=0`

Hence the two lines are perpendicular to each other.