Problem:
If $$l_1^2 + m_1^2 + n_1^2 = l_2^2 + m_2^2 + n_2^2 = l_3^2 + m_3^2 + n_3^2 = 1$$ and $$l_1l_2 + m_1m_2 + n_1n_2 = l_1l_3 + m_1m_3 + n_1n_3 = l_2l_3 + m_2m_3 + n_2n_3 = 0$$ then show that, $$l_1^2 + l_2^2 + l_3^2 = m_1^2 + m_2^2 + m_3^2 = n_1^2 + n_2^2 + n_3^2 = 1$$ and, $$l_1m_1 + l_2m_2 + l_3m_3 = l_1n_1 + l_2n_2 + l_3n_3 = m_1n_1 + m_2n_2 + m_3n_3 = 0$$
My attempt:
For $p = 1, 2, 3$ let, $\vec{u}_p = l_p \hat{\imath} + m_p \hat{\jmath} + n_p \hat{k}$.
Then from the given conditions it is seen that the vectors $\vec{u}_1, \vec{u}_2$ and $\vec{u}_3$ are mutually perpendicular unit vectors.
I was stumped how to proceed after this, so I thought of handling the easier 2-D version of the problem, i.e, letting $n_p = 0$. The given theorems were easily proved since the rectangular components of a unit vector in the x-y are just $\cos \alpha, \sin \alpha$ if $\alpha$ is the angle made with the x-axis. I realized that this property is due to the fact that if the vector makes $\alpha, \beta$ with the x and y axes, then $\cos^2 \alpha + \cos^2 \beta = 1$. Since the two vectors are mutually perpendicular, so if one makes $\alpha$ with x-axis, then the other must make $\dfrac{\pi}{2} - \alpha$. Using this the two theorems to be proven follow.
I know that this property generalizes to 3 dimensions as the sum of squares of direction cosines is 1, but I don't know how to apply this tho solve the problem.
Can somebody give me a hint on how to proceed?
A = $\begin{bmatrix} l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3 \end{bmatrix}$
$AA^T = \begin{bmatrix} l_1^2+m_1^2+n_1^2&l_1l_2 + m_1m_2+n_1n_2& l_1l_3+m_1m_3+n_1n_3\\ l_2l_1+m_2m_2+n_2n_1& l_2^2+m_2^3 + n_2^n &l_2l_3 + m_2m_3+ n_2n_3\\ l_3l_1+m_3m_1+n_3n_1& l_3l_2 + m_3m_2 + n_3n_2&l_3^2+m_3^2 + n_3^2 \end{bmatrix} = \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
i.e. A is othonormal.
if $AA^T = I$ then $A^T = A^{-1}$ and $A^TA = 1$