Let `l_i,m_i,n_i; i=1,2,3` be the direction cosines of three mutually perpendicular vectors in space. Show that `AA'=I_3` where `A=[[l_1,m_1,n_1] , [l_2,m_2,n_2] , [l_3,m_3,n_3]]`

Explanation

Given that `l_{i}, m_{i}, n_{i} ; i=1,2,3` be the direction cosines of three mutually perpendicular vectors in space.

Therefore, `l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0`,


`l_{1} l_{3}+m_{1} m_{3}+n_{1} n_{3}=0` and

`l_{2} l_{3}+m_{2} m_{3}+n_{2} n_{3}=0` .

( `because` sum of product of direction cosines of two perpendicular vector is zero)

Since, `l_{i}, m_{i}, n_{i} ; i=1,2,3` are the direction cosines of vectors.

Therefore,

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