Show that the straight lines whose direction cosines are given b

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110. 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 straight u squared over straight a plus straight v squared over straight b plus straight w squared over straight c equals 0.


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),  straight n space equals space minus fraction numerator straight u space straight l space plus space straight v space straight m over denominator straight w end fraction
Putting this value of n in (2), we get,
al squared plus bm squared plus straight c open parentheses negative fraction numerator ul plus vm over denominator straight w end fraction close parentheses squared space equals space 0
or a wl2 + b w2 m2 + c (ul2 + 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,
left parenthesis straight a space straight w squared plus space straight c space straight u squared right parenthesis space straight l squared over straight m squared plus space 2 space straight c space straight u space straight v space straight l over straight m space plus space left parenthesis straight b space straight w squared plus space straight c space straight v squared right parenthesis space equals space 0                 ...(3)
which is a quadratic in straight l over straight m.
Let l1, m1, n; l2, m2, n2 be the direction-cosines of the two lines. Then straight l subscript 1 over straight m subscript 1 comma space straight l subscript 2 over straight m subscript 2 are the roots of the equation (3).
(i) The lines will be perpendicular when 
       l1 l2 + m1 m2 + n1 n2 = 0    ....(4)
From (3), straight l subscript 1 over straight m subscript 1. space straight l subscript 2 over straight m subscript 2 space equals space fraction numerator straight b space straight w squared plus space straight c space straight v squared over denominator straight a space straight w squared plus space straight c space straight u squared end fraction

therefore space space space space space fraction numerator straight l subscript 1 space straight l subscript 2 over denominator bw squared plus straight c space straight v squared end fraction space equals space fraction numerator straight m subscript 1 space straight m subscript 2 over denominator straight c space straight u squared space plus space straight a space straight w squared end fraction space equals space fraction numerator straight n subscript 1 space straight n subscript 2 over denominator av squared plus bu squared end fraction space equals space straight k space left parenthesis say right parenthesis space space space left parenthesis By space symmetry right parenthesis

∴  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 straight l subscript 1 over straight l subscript 2 space equals space straight m subscript 1 over straight m subscript 2 space space or space space space straight l subscript 1 over straight m subscript 1 space equals space straight l subscript 2 over straight m subscript 2
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 straight w squared over straight c plus straight v squared over straight b plus straight u squared over straight a space equals space 0 space space space space space space space space space space space space space space space space space left square bracket Dividing space by space straight a space straight b space straight c right square bracket

i.e., if straight u squared over straight a plus straight v squared over straight b plus straight w squared over straight c equals 0


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