Let l_i, m_i and n_i; i = 1, 2, 3 be the direction cosines of three mutually perpendicular vectors in space. Show that AA’ = I₃, where A =

Let the magnitude of three vectors be r₁, r₂ and r₃ respectively.

Then, three perpendicular vectors will be

As, vectors are mutually perpendicular, dot product of any two vectors will be zero

v₁. v₂ = l₁r₁l₂r₂ + m₁r₁m₂r₂ + n₁r₁n₂r₂ = 0

⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0 …. [1]

Similarly,

l₂l₃ + m₂m₃ + n₂n₃ = 0 …. [2]

l₁l₃ + m₁m₃ + n₁n₃ = 0 …. [3]

Now, If A = [a_ij]_{m × n} then A’ = [a_ji]_{n × m}

Therefore,

Now, as l_i, m_i and n_i are direction cosiner

⇒ l_i² + m_i² + n_i² = 1 …. [4]

Hence, putting values from [1], [2], [3] and [4] we have

Hence Proved!