# cross product

• Feb 2nd 2007, 06:35 AM
Jenny20
cross product
Can you explain to me why

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

where i , j, k are unit vectors.

Thank you very much.
• Feb 2nd 2007, 07:46 AM
earboth
Quote:

Originally Posted by Jenny20
Can you explain to me why

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

where i , j, k are unit vectors.
Thank you very much.

Hello, Jenny,

I only know the cross product with vectors. To use vectors I re-write your problem:

v_1 = (sin(t), cos(t), 0)

v_2 = (cos(t), -sin(t), 0)

Now calculate the cross-product:

$\begin{array}{|ccc|}i&j&k \\ \sin(t) & \cos(t) & 0 \\ \cos(t) & -\sin(t) &0 \end{array}$ $=(0, 0, \sin(t) \cdot (-\sin(t))-\cos(t) \cdot \cos(t))$

Maybe this helps a little bit further.

EB
• Feb 2nd 2007, 07:53 AM
Soroban
Hello, Jenny!

Quote:

Can you explain to me why

$\left[(\sin t)\vec{i} + (\cos t)\vec{j}\right] \times \left[(\cos t)\vec{i} - (\sin t)\vec{j}\right] \:=\: \left(\text{-}\sin^2t - \cos^2t\right)\vec{k}$ ?

Do you know how to find a cross product?

We have two vectors: . $\vec{u}\:=\:\langle \sin t,\,\cos t,\,0\rangle$ and $\vec{v}\:=\:\langle\cos t,\,\text{-}\sin t,\,0\rangle$

. . Then: . $\vec{u} \times \vec{v}\;=\;\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \\ \sin t & \cos t & 0 \\ \cos t & \text{-}\sin t & 0 \end{vmatrix}$

Now crank it out: . $\vec{i}(\cos t\cdot0 + \sin t\cdot0) - \vec{j}(\sin t\cdot0 - \cos t\cdot0) + \vec{k}(-\sin^2t - \cos^2t)$

Got it?