?Let \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle\) and \(\mathbf{v}=\left\langle

Chapter 2, Problem 227

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Let \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle\) and \(\mathbf{v}=\left\langle v_{1}, v_{2}\right\rangle\) be two-dimensional vectors. The cross product of vectors u and v is not defined. However, if the vectors are regarded as the three-dimensional vectors \(\tilde{\mathbf{u}}=\left\langle u_{1}, u_{2}, 0\right\rangle\) and \(\tilde{\mathbf{v}}=\left\langle v_{1}, v_{2}, 0\right\rangle\) , respectively, then, in this case, we can define the cross product of \(\tilde{\mathbf{u}}\) and \(\tilde{\mathbf{v}}\). In particular, in determinant notation, the cross product of \(\tilde{\mathbf{u}}\) and \(\tilde{\mathbf{v}}\) is given by

\(\tilde{\mathbf{u}} \times \tilde{\mathbf{v}}=\left|\begin{array}{lll}\mathbf{i} & \mathbf{j} & \mathbf{k} \\u_{1} & u_{2} & 0 \\v_{1} & v_{2} & 0\end{array}\right|\)

Use this result to compute \((\mathbf{i} \cos \theta+\mathbf{j} \sin \theta) \times(\mathbf{i} \sin \theta-\mathbf{j} \cos \theta)\), where \(\theta\) is a real number.

Text Transcription:

u = langle u_1, u_2 rangle

v = langle v_1, v_2 rangle

tilde u = langle u_1, u_2, 0 rangle

tilde v = angle v_1, v_2, 0 rangle

tilde u

tilde v

tilde u times tilde v = |i j k over u_1 u_2 0 over v_1 v_2 0|

i cos theta + j sin theta) times (i sin theta - j cos theta)

theta