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#

1 Answer
Mar 12, 2017

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.