What is $\left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$?

Question

I did

\begin{align} & \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2} \\[8pt] = {} & \frac{2t^2i-4tj-2t(t^2+2)k}{(t^2+2)^2} \\[8pt] = {} & \frac{2t^2i-4tj-2t^3k+4tk}{(t^2+2)^2} \\[8pt] = {} & \frac{-4tj+4tk}{(t^2+2)^2} \end{align}

Is this true?

Answer 1

$(\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k) \times \frac{2ti-2tj-(t^2+2)k}{t^2+2}$

$\frac {1}{(t^2 + 2)^2} ((-2(t^2+2)+4t^2) \mathbf i + (4t^2+t^2(t^2+2))\mathbf j + (-2t^3-4t) \mathbf k)$

$\frac {1}{(t^2 + 2)^2} ((2t^2 - 4) \mathbf i + (t^4+6t^2) \mathbf j -(2t^3+4t) \mathbf k)$

$\frac {2t^2 - 4}{(t^2 +2)^2} \mathbf i + \frac {t^4+6t^2}{(t^2 + 2)^2} \mathbf j - \frac {2t^3+4t}{(t^2 + 2)^2} \mathbf k$

Answer 2

\begin{align} i\times j & = k = -j\times i \\ j\times k & = i = -k\times j \\ k\times i & = j = - i\times k \\ i \times i & = j \times j = k\times k = 0 \end{align}