Question
Show that the straight lines whose direction cosines are given by the equations uI + vm + wn = 0, a I2 + b m2 + cn2 = 0 are
(i) perpendicular if u2 (b + c) + v2 (c + a) + w2 (a + b) = 0
(ii) parallel if 
Solution
The direction-cosines of the two lines are given by the equations
u l + v m + w n = 0 ....(1)
and a I2 + b m2 + c n2 = 0 ....(2)
from (1),
Putting this value of n in (2), we get,
or a w2 l2 + b w2 m2 + c (u2 l2 + v2 m2 + 2 u v l m) = 0
or (a w2 + c u2) I2 + 2 c u v I m + (b w2 + c v2) m2 = 0
Dividing both sides by m2, we get,
...(3)
which is a quadratic in
Let l1, m1, n1 ; l2, m2, n2 be the direction-cosines of the two lines. Then
are the roots of the equation (3).
(i) The lines will be perpendicular when
l1 l2 + m1 m2 + n1 n2 = 0 ....(4)
From (3),
u l + v m + w n = 0 ....(1)
and a I2 + b m2 + c n2 = 0 ....(2)
from (1),
Putting this value of n in (2), we get,
or a w2 l2 + b w2 m2 + c (u2 l2 + v2 m2 + 2 u v l m) = 0
or (a w2 + c u2) I2 + 2 c u v I m + (b w2 + c v2) m2 = 0
Dividing both sides by m2, we get,
...(3)which is a quadratic in
Let l1, m1, n1 ; l2, m2, n2 be the direction-cosines of the two lines. Then
are the roots of the equation (3).(i) The lines will be perpendicular when
l1 l2 + m1 m2 + n1 n2 = 0 ....(4)
From (3),
∴ l1 l2 = k (b w2 + c v2), m1 m2 = k (c u2 + a w2), n1 n2 = k (a v2 + b u2)
∴ l1 l2 + m1 m2 + n1 n2 = k [b w2'+ cv2 + c u2 + a w'2 + a v2 + b u2]
∴ lines are perpendicular
If k (b w2 + cv2 + c u2 + a w2 + a v2 + b u2] = 0 [∵ of (4)]
i.e., if b w2 + c v2 + c u2 + a w2 + a v2 + b u2 = 0
i.e., if u2 (b + c) + v2 (c + a) + w2 (a + b) = 0
(ii) The lines are parallel
if l1 = l2, m1 = m2, n1 = n2
i.e., if 
i.e., if equation (3) has equal roots
i.e., if disc = 0
i.e., if 4 c2 u2 v2 – 4 (a w2 + c u2) (b w2 + c v2) = 0
i.e., lf c2 u2 v2 – a bw4 – acv2 w2 –bcu2 W – c2 u2 v2= 0
i.e., if – a b w4 – a c v2 w2 – b c u2 w2 = 0
i.e., if a b w2 + a c v2 + b c u2 = 0 [Dividing by – w2]
i.e, if 
i.e., if 
