Selftest: Cross product

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Exercises for chapter Vector algebra | Article: Cross product

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1. Please mark the right transforms:

	$\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}} \upuparrows\vec{\mathbf{b}}$
	$\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}}\bot\vec{\mathbf{b}}$
→	The magnitude of the cross product results from the following equation: $\vec{\mathbf{a}}\times\vec{\mathbf{b}} =\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right| = ab\sin\alpha$. In this case the angle is 90° and so the magnitude is |ab|.
	$\vec{\mathbf{a}}\times\vec{\mathbf{b}}= \vec{\mathbf{e}}_cab \text{ for } \vec{\mathbf{a}}\uparrow\downarrow\vec{\mathbf{b}}$
→	The magnitude of the cross product results from the following equation: $\vec{\mathbf{a}}\times\vec{\mathbf{b}}=\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right|= ab\sin\alpha$. Here the angle is 180° and the related sine is zero. So the result is 0.

2. Please mark the right transforms: (x,y,z are coordinate axes, a and b are arbitrary)

	$\vec{\mathbf{b}} \times \vec{\mathbf{a}}=-(\vec{\mathbf{a}}\times \vec{\mathbf{b}})$
	$\vec{\mathbf{y}}\times\vec{\mathbf{x}}=-\vec{\mathbf{z}}$
	The cross product is commutative.
→	The first answer implies that the cross product is not commutative. This can easily be proved by the right-hand-rule. Further information: see Cross product

3. Please solve the following exercise:

$\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}=$$\vec{\mathbf{e}}_x+$$\vec{\mathbf{e}}_y+$$\vec{\mathbf{e}}_z$

→ To compute the cross product the component representation $\vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3$ is used.

4. Please solve the following exercise:

$\begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}=$$\vec{\mathbf{e}}_x+$$\vec{\mathbf{e}}_y+$$\vec{\mathbf{e}}_z$

→ To compute the cross product the component representation $\vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3$ is used.

5. Please solve the following exercise:

$\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=$$\vec{\mathbf{e}}_x+$$\vec{\mathbf{e}}_y+$$\vec{\mathbf{e}}_z$

→ To compute the cross product the component representation $\vec{\mathbf{a}} \times \vec{\mathbf{b}} =(a_2 b_3 - a_3 b_2) \vec{\mathbf{e}}_1 + (a_3 b_1 - a_1 b_3) \vec{\mathbf{e}}_2 + (a_1 b_2 - a_2 b_1) \vec{\mathbf{e}}_3$ is used.

6. The following vectors are given.

Please mark the correct statements sonsidering the the given vectors.

	The vector $\vec{\mathbf{a}}\times\vec{\mathbf{b}}$ is perpendicular to the plane that is spanned by the vectors $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$.
	The magnitude of the cross product equals the area of the parallelogram that is spanned by the vectors $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$.
	The vectors $\vec{\mathbf{a}}$, $\vec{\mathbf{b}}$ and $\vec{\mathbf{a}}\times\vec{\mathbf{b}}$ form a rectangular coordinate system. So they are arranged as thumb, middle finger and forefinger of the right hand.

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