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

Question

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

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Answer 1 · 2017-03-12T09:33:26

The two lines are perpendicular to each other.

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.