Difference between revisions of "Selftest: Cross product"
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− | {{ExerciseNavigation|previous=[[Selftest:Dot product|Dot product]]| | + | {{ExerciseNavigation|previous=[[Selftest: Dot product|Dot product]]|chapter=[[Vector algebra]]|article=[[Cross product]]|next=[[Selftest: Introduction to vector algebra|Introduction to vector algebra]]}} |
− | <quiz> | + | <quiz display=simple> |
{'''Please mark the right transforms:''' } | {'''Please mark the right transforms:''' } | ||
+ <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}} \upuparrows\vec{\mathbf{b}}</math> | + <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}= 0 \text{ for } \vec{\mathbf{a}} \upuparrows\vec{\mathbf{b}}</math> | ||
Line 10: | Line 10: | ||
||The magnitude of the cross product results from the following equation: <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}=\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right|= ab\sin\alpha</math>. Here the angle is 180° and the related sine is zero. So the result is 0. | ||The magnitude of the cross product results from the following equation: <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}=\vec{\mathbf{c}}\text{ with}\left|\vec{\mathbf{c}}\right|= ab\sin\alpha</math>. Here the angle is 180° and the related sine is zero. So the result is 0. | ||
− | {''' | + | {'''Please mark the right transforms: (x,y,z are coordinate axes, a and b are arbitrary)''' } |
+ <math>\vec{\mathbf{b}} \times \vec{\mathbf{a}}=-(\vec{\mathbf{a}}\times \vec{\mathbf{b}})</math> | + <math>\vec{\mathbf{b}} \times \vec{\mathbf{a}}=-(\vec{\mathbf{a}}\times \vec{\mathbf{b}})</math> | ||
+ <math>\vec{\mathbf{y}}\times\vec{\mathbf{x}}=-\vec{\mathbf{z}} </math> | + <math>\vec{\mathbf{y}}\times\vec{\mathbf{x}}=-\vec{\mathbf{z}} </math> | ||
− | - | + | -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]] |
− | {''' | + | {'''Please solve the following exercise:''' |
| type="{}" } | | type="{}" } | ||
<math>\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}=</math>{ 2 }<math>\vec{\mathbf{e}}_x+</math>{ 2 }<math>\vec{\mathbf{e}}_y+</math>{ -1 }<math>\vec{\mathbf{e}}_z</math> | <math>\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}=</math>{ 2 }<math>\vec{\mathbf{e}}_x+</math>{ 2 }<math>\vec{\mathbf{e}}_y+</math>{ -1 }<math>\vec{\mathbf{e}}_z</math> | ||
− | || | + | ||To compute the cross product the component representation <math>\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 </math> is used. |
− | {''' | + | {'''Please solve the following exercise:''' |
| type="{}" } | | type="{}" } | ||
<math>\begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}=</math>{ 1 }<math>\vec{\mathbf{e}}_x+</math>{ -2 }<math>\vec{\mathbf{e}}_y+</math>{ 3 }<math>\vec{\mathbf{e}}_z</math> | <math>\begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} -2 \\ -1 \\ 0 \end{pmatrix}=</math>{ 1 }<math>\vec{\mathbf{e}}_x+</math>{ -2 }<math>\vec{\mathbf{e}}_y+</math>{ 3 }<math>\vec{\mathbf{e}}_z</math> | ||
− | || | + | ||To compute the cross product the component representation <math>\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 </math> is used. |
− | {''' | + | {'''Please solve the following exercise:''' |
| type="{}" } | | type="{}" } | ||
<math>\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=</math>{ -1 }<math>\vec{\mathbf{e}}_x+</math>{ 0.0 }<math>\vec{\mathbf{e}}_y+</math>{ 0.0 }<math>\vec{\mathbf{e}}_z</math> | <math>\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=</math>{ -1 }<math>\vec{\mathbf{e}}_x+</math>{ 0.0 }<math>\vec{\mathbf{e}}_y+</math>{ 0.0 }<math>\vec{\mathbf{e}}_z</math> | ||
− | || | + | ||To compute the cross product the component representation <math>\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 </math> is used. |
− | { ''' | + | { '''The following vectors are given.''' |
− | [[ | + | [[File:Vectoralgebra_crossproduct.jpg|300px]] |
− | + | Please mark the correct statements sonsidering the the given vectors. | |
} | } | ||
− | + | + | + The vector <math>\vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> is perpendicular to the plane that is spanned by the vectors <math>\vec{\mathbf{a}}</math> and <math>\vec{\mathbf{b}}</math>. |
− | + | + The magnitude of the cross product equals the area of the parallelogram that is spanned by the vectors <math>\vec{\mathbf{a}}</math> and <math>\vec{\mathbf{b}}</math>. | |
− | + | + The vectors <math>\vec{\mathbf{a}}</math>, <math>\vec{\mathbf{b}}</math> and <math> \vec{\mathbf{a}}\times\vec{\mathbf{b}}</math> form a rectangular coordinate system. So they are arranged as thumb, middle finger and forefinger of the right hand. | |
− | + | ||
</quiz> | </quiz> | ||
[[Category:Selftest]] | [[Category:Selftest]] | ||
[[Category:Vectors]] | [[Category:Vectors]] |
Latest revision as of 10:18, 25 September 2014
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