Difference between revisions of "Selftest: Cross product"
From Robotics
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||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> | ||
− | || | + | ||There are two possibilities to compute the cross product. Either 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 or the angle between the vectors: <math>\vec{\mathbf{c}} = ab\sin\alpha</math> |
− | {''' | + | {'''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> | ||
− | || | + | ||There are two possibilities to compute the cross product. Either 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 or the angle between the vectors: <math>\vec{\mathbf{c}} = ab\sin\alpha</math> |
− | {''' | + | {'''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> | ||
− | || | + | ||There are two possibilities to compute the cross product. Either 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 or the angle between the vectors: <math>\vec{\mathbf{c}} = ab\sin\alpha</math> |
Revision as of 16:50, 23 May 2014
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