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If l 1 , m 1 , n 1 ; l 2 , m 2 , n 2 , and   l 3 , m 3 , n 3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l 1 + l 2 + l 3 , m 1 + m 2 + m 3 , n 1 + n 2 + n 3 and  makes equal angles with them.

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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a long answer type question as classified in NCERT Exemplar

    L e t     a → ,   b → ,   c →     a n d     d →     a r e     s u c h     t h a t                                 a → = l 1 i ^ + m 1 j ^ + n 1 k ^                                 b → = l 2 i ^ + m 2 j ^ + n 2 k ^                                 c → = l 3 i ^ + m 3 j ^ + n 3 k ^ a n d               d → = ( l 1 + l 2 + l 3 ) i ^ + ( m 1 + m 2 + m 3 ) j ^ + ( n 1 + n 2 + n 3 ) k ^ Since  the  given  d'cosines  are  mutually  perpendicular  then                               l 1 l 2 + m 1 m 2 + n 1 n 2 = 0                               l 2 l 3 + m 2 m 3 + n 2 n 3 = 0                               l 1 l 3 + m 1 m 3 + n 1 n 3 = 0 L e t     α ,   β     a n d     γ     b e     t h e     a n g l e s     b e t w e e n     a →     a n d     d → ,     b →     a n d     d → ,     c →     a n d     d →     r e s p e c t i v e l y . ∴                       c o s α = l 1 ( l 1 + l 2 + l 3 ) + m 1 ( m 1 + m 2 + m 3 ) + n 1 ( n 1 + n 2 + n 3 )                                                     = l 1 2 + l 1 l 2 + l 1 l 3 + m 1 2 + m 1 m 2 + m 1 m 3 + n 1 2 + n 1 n 2 + n 1 n 3                                                     = ( l 1 2 + m 1 2 + n 1 2 ) + ( l 1 l 2 + m 1 m 2 + n 1 n 2 ) + ( l 1 l 3 + m 1 m 3 + n 1 n 3 )                                                     = 1 + 0 + 0 = 1 ∴                       c o s β = l 2 ( l 1 + l 2 + l 3 ) + m 2 ( m 1 + m 2 + m 3 ) + n 2 ( n 1 + n 2 + n 3 )                                                     = l 1 l 2 + l 2 2 + l 2 l 3 + m 1 m 2 + m 2 2 + m 2 m 3 + n 1 n 2 + n 2 2 + n 2 n 3                                                     = ( l 2 2 + m 2 2 + n 2 2 ) + ( l 1 l 2 + m 1 m 2 + n 1 n 2 ) + ( l 2 l 3 + m 2 m 3 + n 2 n 3 )                                                     = 1 + 0 + 0 = 1 S i m i l a r l y , ∴                       c o s γ = l 3 ( l 1 + l 2 + l 3 ) + m 3 ( m 1 + m 2 + m 3 ) + n 3 ( n 1 + n 2 + n 3 )                                                     = l 1 l 3 + l 2 l 3 + l 3 2 + m 1 m 3 + m 2 m 3 + m 3 2 + n 1 n 3 + n 2 n 3 + n 3 2                                                     = ( l 3 2 + m 3 2 + n 3 2 ) + ( l 1 l 3 + m 1 m 3 + n 1 n 3 ) + ( l 2 l 3 + m 2 m 3 + n 2 n 3 )                                                     = 1 + 0 + 0 = 1 ∴                       c o s α = c o s β = c o s γ = 1         ⇒ α = β = γ     w h i c h     i s     t h e     r e q u i r e d     r e s u l t .

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