The direction cosines of a line satisfy the relations `lambda(l+m)=n` and `m n+n l+l m=0.` The value of `lambda,` for which the two lines are perpendicular to each other, is

Options

A. 1
B. 2
C. `1//2`
D. none of these

Answer

b

Explanation

Eliminating `n`, we get
`" "lamda(l+m)^(2)+lm=0`
`rArr" "(lamdal^(2))/(m^(2))+ (2lamda+1)(l)/(m) + lamda=0`
`rArr" "(l_1l_2)/(m_(1)m_(2))=1" "` (product of roots `(l_1)/(m_1) and (l_2)/(m_2)` )
where `l_1//m_1 and l_2//m_2` are the roots of this equation, further eliminating m, we get
`" "lamdal^(2)-ln-n^(2)=0`
`rArr" "(l_(1)l_(2))/(n_(1)n_(2))=-(1)/(lamda)`
Since the lines with direction cosines `(l_1, m_1, n_1) and (l_2, m_2, n_2)` are perpendicular, we have
`" "l_1l_2+ m_1m_2 + n_1n_2 = 0`
or `" "1+1-lamda=0`
or `" "lamda=2`