Selftest: Cross product

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Exercises for chapter {{{chapter}}} | Article: Vector algebra

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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 equation of the magnitude of the cross product not only includes the dot product but also the sine of the angle between the vectors. So this answer is wrong.
	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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