another cross product

**Jenny20** · February 5th 2007, 08:25 AM

(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.

**topsquark** · February 5th 2007, 08:40 AM

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.

$<tsin(t), t cos(t), 0> \times <sin(t) + tcos(t), cos(t) - tsin(t), 0>$

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

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

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