How to prove that using law of cosines? I'm reading a book about Cartesian Tensors. The author said that is possible to obtain the following, applying the law of cosines.
\begin{equation}
\sum_{k=1}^{3}\lambda_{ik}\lambda_{jk} = 0\qquad i,j = 1,2,3 \qquad\ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\label{eq1}
\end{equation}
where $\lambda$ is a rotation matrix. I don't know how to obtain (1) using just the law of cosines. I can obtain (1) by applying three consecutive rotations in regarding to $x_1$, $x_2$ and $x_3$ axes, yielding a rotation matrix
$$R = R(\alpha)R(\beta)R(\gamma) =
 \left(
 \begin{array}{ccc}
  \cos (\beta ) \cos (\gamma ) & \cos (\beta ) \sin (\gamma ) & \sin (\beta ) \\
  -\sin (\alpha ) \sin (\beta )\cos (\gamma )-\cos (\alpha ) \sin (\gamma ) & \cos (\alpha ) \cos (\gamma )-\sin (\alpha ) \sin (\beta ) \sin (\gamma ) & \sin (\alpha ) \cos (\beta ) \\
  \sin (\alpha ) \sin (\gamma )-\cos (\alpha ) \sin (\beta ) \cos (\gamma ) & -\sin (\alpha )\cos (\gamma )-\cos (\alpha ) \sin (\beta ) \sin (\gamma ) & \cos (\alpha ) \cos (\beta ) \\
 \end{array}
 \right)
$$
then applying (1) above I can obtain the results equal to zero. But I'm curious to know how can I get it using a geometrical approach, e.g. law of cosines or another geometrical approach.
 A: The approach I find most intuitive is the following: the key point is that a rotation is an isometric (equivalently "orthogonal") linear transformation: the length of a vector $x$ is equal to the length of the transformed vector $\lambda x$.
Consider any $x = (x_1,x_2,x_3) \in \Bbb R^3$. We have
\begin{align}
\|\lambda x\|^2 &= 
\sum_{i=1}^3 (\lambda x)_i^2 = \sum_{i=1}^3 \left(\sum_{j=1}^3 \lambda_{ij} x_j\right)^2
\\ & = \sum_{i=1}^3 \left(\sum_{j=1}^3 \lambda_{ij} x_j\right)\left(\sum_{k=1}^3 \lambda_{ik} x_k\right)
= \sum_{i,j,k=1}^3 \lambda_{ij} \lambda_{ik} x_j x_k.
\end{align}
Let $M$ denote the matrix with $M_{jk} = \sum_{i} \lambda_{ij}\lambda_{ik}$ (that is, $M = \lambda^T\lambda$). Note that this matrix is symmetric, which is to say that $M_{jk} = M_{kj}$. From the above, we have $\|\lambda x\|^2 = \sum_{jk} M_{jk} x_jx_k$. On the other hand, from the isometric property we have
$$
\|\lambda x\|^2 = \|x\|^3 = \sum_{j}x_j^2 = \sum_{jk} \delta_{jk} x_j x_k,
$$
where $\delta_{jk}$ denote the Kronecker delta. That is,
$$
\sum_{jk} M_{jk}x_jx_k = \sum_{jk}\delta_{jk} x_jx_k.
$$
Now, argue that because $M$ is symmetric this holds for all choices of $x \in \Bbb R^3$, it must be that $M_{jk} = \delta_{jk}$. That is, we have
$$
\lambda^T \lambda = I \implies \lambda^T = \lambda^{-1} \implies \lambda\lambda^T = I \implies\\
\sum_k \lambda_{ij}\lambda_{jk} = \delta_{ij},
$$
which is what we wanted to show.
