Cross Product
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
The em vector product of two vectors veca leftmatrixa_xa_ya_zmatrixright textrm and vecb leftmatrixb_xb_yb_zmatrixright is given by leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright abcliste abc Show that for veca along the x axis and vecb along the y axis vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb in the xy plane vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb vecacrossvecb is perpicular to both veca and vecb. abcliste
Solution:
abcliste abc The vectors veca and vecb can be written as veca leftmatrixamatrixright textrm and vecb leftmatrixbmatrixright It follows for the cross product vecacrossvecb leftmatrix - b - a a b - matrixright leftmatrixa bmatrixright This is obviously a vector along the z axis. abc The vectors veca and vecb can be written as veca leftmatrixa_xa_ymatrixright textrm and vecb leftmatrixb_xb_ymatrixright It follows for the cross product vecacrossvecb leftmatrixa_y - b_y b_x - a_x a_x b_y - a_y b_xmatrixright leftmatrixa_x b_y - a_y b_xmatrixright This is obviously a vector along the z axis. abc Two vectors are perpicular to each other if their dot product is equal to zero. The dot product of vecacrossvecb and veca is leftvecacrossvecbright veca leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright leftmatrixa_xa_ya_zmatrixright a_x a_y b_z-a_x a_z b_y+a_y a_z b_x nonumber & quad - a_x a_y b_z +a_x a_z b_y - a_y a_z b_x a_y a_z-a_y a_zb_x nonumber & quad +a_x a_z-a_x a_zb_y nonumber & quad +a_x a_y-a_x a_yb_z The verification for vecb works in the same way. abcliste
The em vector product of two vectors veca leftmatrixa_xa_ya_zmatrixright textrm and vecb leftmatrixb_xb_yb_zmatrixright is given by leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright abcliste abc Show that for veca along the x axis and vecb along the y axis vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb in the xy plane vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb vecacrossvecb is perpicular to both veca and vecb. abcliste
Solution:
abcliste abc The vectors veca and vecb can be written as veca leftmatrixamatrixright textrm and vecb leftmatrixbmatrixright It follows for the cross product vecacrossvecb leftmatrix - b - a a b - matrixright leftmatrixa bmatrixright This is obviously a vector along the z axis. abc The vectors veca and vecb can be written as veca leftmatrixa_xa_ymatrixright textrm and vecb leftmatrixb_xb_ymatrixright It follows for the cross product vecacrossvecb leftmatrixa_y - b_y b_x - a_x a_x b_y - a_y b_xmatrixright leftmatrixa_x b_y - a_y b_xmatrixright This is obviously a vector along the z axis. abc Two vectors are perpicular to each other if their dot product is equal to zero. The dot product of vecacrossvecb and veca is leftvecacrossvecbright veca leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright leftmatrixa_xa_ya_zmatrixright a_x a_y b_z-a_x a_z b_y+a_y a_z b_x nonumber & quad - a_x a_y b_z +a_x a_z b_y - a_y a_z b_x a_y a_z-a_y a_zb_x nonumber & quad +a_x a_z-a_x a_zb_y nonumber & quad +a_x a_y-a_x a_yb_z The verification for vecb works in the same way. abcliste
Meta Information
Exercise:
The em vector product of two vectors veca leftmatrixa_xa_ya_zmatrixright textrm and vecb leftmatrixb_xb_yb_zmatrixright is given by leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright abcliste abc Show that for veca along the x axis and vecb along the y axis vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb in the xy plane vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb vecacrossvecb is perpicular to both veca and vecb. abcliste
Solution:
abcliste abc The vectors veca and vecb can be written as veca leftmatrixamatrixright textrm and vecb leftmatrixbmatrixright It follows for the cross product vecacrossvecb leftmatrix - b - a a b - matrixright leftmatrixa bmatrixright This is obviously a vector along the z axis. abc The vectors veca and vecb can be written as veca leftmatrixa_xa_ymatrixright textrm and vecb leftmatrixb_xb_ymatrixright It follows for the cross product vecacrossvecb leftmatrixa_y - b_y b_x - a_x a_x b_y - a_y b_xmatrixright leftmatrixa_x b_y - a_y b_xmatrixright This is obviously a vector along the z axis. abc Two vectors are perpicular to each other if their dot product is equal to zero. The dot product of vecacrossvecb and veca is leftvecacrossvecbright veca leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright leftmatrixa_xa_ya_zmatrixright a_x a_y b_z-a_x a_z b_y+a_y a_z b_x nonumber & quad - a_x a_y b_z +a_x a_z b_y - a_y a_z b_x a_y a_z-a_y a_zb_x nonumber & quad +a_x a_z-a_x a_zb_y nonumber & quad +a_x a_y-a_x a_yb_z The verification for vecb works in the same way. abcliste
The em vector product of two vectors veca leftmatrixa_xa_ya_zmatrixright textrm and vecb leftmatrixb_xb_yb_zmatrixright is given by leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright abcliste abc Show that for veca along the x axis and vecb along the y axis vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb in the xy plane vecacrossvecb is a vector along the z axis. abc Show that for any two vectors veca and vecb vecacrossvecb is perpicular to both veca and vecb. abcliste
Solution:
abcliste abc The vectors veca and vecb can be written as veca leftmatrixamatrixright textrm and vecb leftmatrixbmatrixright It follows for the cross product vecacrossvecb leftmatrix - b - a a b - matrixright leftmatrixa bmatrixright This is obviously a vector along the z axis. abc The vectors veca and vecb can be written as veca leftmatrixa_xa_ymatrixright textrm and vecb leftmatrixb_xb_ymatrixright It follows for the cross product vecacrossvecb leftmatrixa_y - b_y b_x - a_x a_x b_y - a_y b_xmatrixright leftmatrixa_x b_y - a_y b_xmatrixright This is obviously a vector along the z axis. abc Two vectors are perpicular to each other if their dot product is equal to zero. The dot product of vecacrossvecb and veca is leftvecacrossvecbright veca leftmatrixa_y b_z - a_z b_ya_z b_x - a_x b_za_x b_y - a_y b_xmatrixright leftmatrixa_xa_ya_zmatrixright a_x a_y b_z-a_x a_z b_y+a_y a_z b_x nonumber & quad - a_x a_y b_z +a_x a_z b_y - a_y a_z b_x a_y a_z-a_y a_zb_x nonumber & quad +a_x a_z-a_x a_zb_y nonumber & quad +a_x a_y-a_x a_yb_z The verification for vecb works in the same way. abcliste
Contained in these collections:
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Force Vectors by by
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Magnetic Forces by by