# Solve this

Question:

If $I_{i}, m_{i}, n_{j}, i=1,2,3$ denote the direction cosines of three mutually perpendicular vectors in space, prove that $A A^{T}=I$,$=l$, where $A=$$\left[\begin{array}{lll}l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3}\end{array}\right]$

Solution:

Given.

$\left(l_{1}, m_{1}, n_{1}\right),\left(l_{2}, m_{2}, n_{2}\right),\left(l_{3}, m_{3}, n_{3}\right)$ are the direction cosines of three mutually perpendicular vectors in space.

$\left.\begin{array}{l}l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1 \\ l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=1 \\ l_{3}^{2}+m_{3}^{2}+n_{3}^{2}=1\end{array}\right\}$ ....(1)

$\left.\begin{array}{l}l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0 \\ l_{2} l_{3}+m_{2} m_{3}+n_{2} n_{3}=0 \\ l_{3} l_{1}+m_{3} m_{1}+n_{3} n_{1}=0\end{array}\right\}$      ...(2)

Let $A=\left[\begin{array}{lll}l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3}\end{array}\right]$

$\Rightarrow A^{T}=\left|\begin{array}{ccc}l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3}\end{array}\right|$

$A A^{T}=\left[\begin{array}{lll}l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \\ l_{3} & m_{3} & n_{3}\end{array}\right]\left[\begin{array}{ccc}l_{1} & l_{2} & l_{3} \\ m_{1} & m_{2} & m_{3} \\ n_{1} & n_{2} & n_{3}\end{array}\right]$

$\Rightarrow A A^{T}=\left[\begin{array}{ccc}l_{1}{ }^{2}+m_{1}{ }^{2}+n_{1}{ }^{2} & l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2} & l_{3} l_{1}+m_{3} m_{1}+n_{3} n_{1} \\ l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2} & l_{2}{ }^{2}+m_{2}{ }^{2}+n_{2}{ }^{2} & l_{2} l_{3}+m_{2} m_{3}+n_{2} n_{3} \\ l_{3} l_{1}+m_{3} m_{1}+n_{3} n_{1} & l_{2} l_{3}+m_{2} m_{3}+n_{2} n_{3} & l_{3}{ }^{2}+m_{3}{ }^{2}+n_{3}{ }^{2}\end{array}\right]$

From (i) and (ii), we get

$A A^{T}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$

Hence proved.