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:
$\displaystyle \begin{array}{|ccc|}i&j&k \\ \sin(t) & \cos(t) & 0 \\ \cos(t) & -\sin(t) &0 \end{array}$$\displaystyle =(0, 0, \sin(t) \cdot (-\sin(t))-\cos(t) \cdot \cos(t))$
Maybe this helps a little bit further.
EB
Hello, Jenny!
Do you know how to find a cross product?Can you explain to me why
$\displaystyle \left[(\sin t)\vec{i} + (\cos t)\vec{j}\right] \times \left[(\cos t)\vec{i} - (\sin t)\vec{j}\right] \:=\: \left(\text{-}\sin^2t - \cos^2t\right)\vec{k}$ ?
We have two vectors: .$\displaystyle \vec{u}\:=\:\langle \sin t,\,\cos t,\,0\rangle$ and $\displaystyle \vec{v}\:=\:\langle\cos t,\,\text{-}\sin t,\,0\rangle$
. . Then: .$\displaystyle \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: .$\displaystyle \vec{i}(\cos t\cdot0 + \sin t\cdot0) - \vec{j}(\sin t\cdot0 - \cos t\cdot0) + \vec{k}(-\sin^2t - \cos^2t) $
Got it?