proving $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation

Question

R.T.P $|| \vec{x} \times \vec{y}||^{2} = ||\vec{x}||^{2}||\vec{y}||^{2} - (\vec{x} \cdot \vec{y})^{2}$ using index notation.

where $\epsilon_{ijk} = \begin{cases} 1 : (i,j,k) \in \lbrace (1,2,3);(2,3,1);(3,1,2) \rbrace \\ -1: (i,j,k) \in \lbrace (1,3,2);(2,1,3);(3,2,1) \rbrace \\ 0: i=j;i=k;j=k \end{cases}$

initially i used the fact that:

$\vec{x} \times \vec{y} = \epsilon_{ijk} \vec{e_{i}} x_{j} y_{k}$

and that

$|| \vec{x} \times \vec{y} || = \epsilon_{ijk} x_{j} y_{k}$

$|| \vec{x} \times \vec{y} ||^{2} = (\epsilon_{ijk} x_{j} y_{k})^{2}$

im unable to split it into 2 terms that is the main issue.

Answer 1

First, observe that the dot product $\vec{x}\cdot\vec{y}$ of two vectors can be written as $$\vec{x}\cdot\vec{y}=\delta_{ij}x_iy_j=x_iy_i,$$ and that the square of the norm $|\vec{x}\rvert^2 = \vec{x}\cdot \vec{x}$. Now, \begin{align}\lvert\vec{x}\times \vec{y}\rvert^2&=(\vec{x}\times \vec{y})\cdot (\vec{x}\times\vec{y})\\ &= (\varepsilon_{ijk}x_jy_k)\:(\varepsilon_{ilm}x_ly_m) \\ &= (\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})\:x_jx_ly_ky_m \\ &= x_jx_jy_ky_k-x_jy_jx_ky_k \\ &=|\vec{x}|^2|\vec{y}|^2-(\vec{x}\cdot\vec{y})^2. \end{align} We have also used the following identity: $$\varepsilon_{ijk}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}.$$

Answer 2

Observe that \begin{equation*} \begin{split}||x||^2||y||^2 -(x\cdot y)^2 & = \left(\sum_{j}x_j^2\right)\left(\sum_{k}y_k^2\right) - \left(\sum_{l}x_ly_l\right)^2 \\ &= \sum_{j\neq k}\left(x^2_jy_k^2 + x_k^2y_j^2-2x_jy_jx_ky_k\right) \\ & = \sum_{j\neq k}(x_jy_k - x_jy_k)^2 \\ & = \sum_{i}\left(\sum_{j}\sum_{k} \epsilon_{ijk}x_jy_k\right)^2 =||x\times y||^2 \end{split} \end{equation*}