Math Help - another cross product

1. another cross product

(tsint i + tcost j) x (( sint + tcost )i + (cost -tsint)j) =-t^2(sin^2 t+ cos^2t)k

Could you please teach me how to get this answer on the RHS? Thank you very much.

2. Originally Posted by Jenny20
(tsint i + tcost j) x (( sint + tcost )i + (cost -tsint)j) =-t^2(sin^2 t+ cos^2t)k

Could you please teach me how to get this answer on the RHS? Thank you very much.
$ \times $

$= \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ tsin(t) & tcos(t) & 0 \\ sin(t) + tcos(t) & cos(t) - tsin(t) & 0 \end{array} \right |$

Expanding the determinant across the top row:
$= [0 \cdot tcos(t) - 0 \cdot (cos(t) - tsin(t)] \hat{i} - [0 \cdot tsint(t) - 0 \cdot (sin(t) + tcos(t)] \hat{j}$ $+ [tsin(t) \cdot (cos(t) - tsin(t)) - tcos(t) \cdot (sin(t) + tcos(t)] \hat{k}$

The $\hat{i}$ and $\hat{j}$ coefficients are obviously 0, so we get:
$= [tsin(t) \cdot (cos(t) - tsin(t)) - tcos(t) \cdot (sin(t) + tcos(t)] \hat{k}$

$= [tsin(t)cos(t) - t^2sin^2(t) - tsin(t)cos(t) - t^2cos^2(t)] \hat{k}$

$= -t^2 \hat{k}$

-Dan