IB Math HL - Cross Product Lesson

00:19:38
https://www.youtube.com/watch?v=wA_OfOQeGnw

Summary

TLDRLa leçon discute du produit vectoriel, également appelé produit croisé, en expliquant comment il est utilisé pour déterminer l'aire d'un parallélogramme formé par deux vecteurs. On passe d'abord par les concepts de base, où l'aire est calculée en utilisant la magnitudes des vecteurs implicants et l'angle entre eux via le sinus. Le produit vectoriel est perpendiculaire aux deux vecteurs entrants, direction déterminée par la règle de la main droite. La formule standard est introduite pour calculer ce produit vectoriel en utilisant les coordonnées des vecteurs impliqués. De plus, diverses propriétés et applications du produit vectoriel sont explorées, comme son rôle dans le calcul des vecteurs unitaires et les produits scalaires triples. Un accent particulier est mis sur l'importante différence de signe entre A x B et B x A, et leur impact sur les questions géométriques et la direction des résultats.

Takeaways

  • 📐 L'aire d'un parallélogramme est donnée par |A| |B| sin(θ).
  • 🔄 Le produit vectoriel de A et B résulte en un vecteur perpendiculaire.
  • ✋ La direction du produit vectoriel est déterminée par la règle de la main droite.
  • 📘 La formule du produit vectoriel utilise i, j, k pour déterminer les composantes.
  • 🔄 A x B = - (B x A), important pour la direction et le sens.
  • 🧮 Les calculs incluent souvent la magnitude pour les vecteurs unitaires.
  • 📏 Le produit scalaire triple génère une valeur scalaire, non vectorielle.
  • 0️⃣ Le produit vectoriel entre deux vecteurs identiques est le vecteur nul.
  • 🧭 Un vecteur unitaire est obtenu en normalisant le vecteur produit.
  • ➗ Le produit vectoriel est distribué à travers les additions de vecteurs.

Timeline

  • 00:00:00 - 00:05:00

    Aujourd'hui, nous étudions le produit vectoriel ou produit croisé, illustré par un parallélogramme. Pour calculer l'aire de celui-ci, nous avons besoin de la hauteur, qui est la magnitude de B multipliée par le sinus de l'angle θ entre les vecteurs A et B. Ainsi, l'aire est égale au produit des magnitudes des vecteurs A et B multipliée par le sinus de θ, ce qui correspond aussi à la magnitude du produit croisé A × B, obtenu par application de la règle de la main droite. Ce produit résulte en un vecteur perpendiculaire à A et B.

  • 00:05:00 - 00:10:00

    Le produit croisé est explicité par exemple entre les vecteurs A et B, composés de coordonnées respectives. Le calcul se fait avec les composantes i, j, k, suivant un processus déterminé par des opérations croisées et soustraction entre les termes des vecteurs. Le produit A × B est obtenu, démontrant que les vecteurs résultants de A × B et B × A sont opposés, soulignant la propriété anti-commutative du produit croisé. Ce produit résulte en un vecteur dont la magnitude peut être calculée pour obtenir des informations telles que l'angle entre A et B.

  • 00:10:00 - 00:19:38

    Enfin, l'application du produit croisé à plusieurs scénarios démontrent son utilité pour calculer des aires de parallélogrammes et les vecteurs unitaires perpendiculaires à un plan défini par trois points. Des propriétés importantes incluent sa distributivité et le produit mixte scalaire. Les vecteurs parallèles aboutissent à un produit croisé nul, et le signe affecte le sens du vecteur résultant. L'importance fondamentale est que le produit croisé génère un vecteur perpendiculaire aux vecteurs d'origine, et sa magnitude correspond à l'aire du parallélogramme formé par ces vecteurs.

Mind Map

Video Q&A

  • Quel est le résultat d'un produit vectoriel en termes de type de résultat?

    Le produit vectoriel résulte en un vecteur.

  • Comment calcule-t-on l'aire d'un parallélogramme en utilisant des vecteurs?

    Un parallélogramme formé par des vecteurs A et B a pour aire le produit des magnitudes de A et B multiplié par le sinus de l'angle entre eux.

  • Que se passe-t-il lorsque l'on effectue le produit vectoriel d'un vecteur par lui-même?

    Le produit vectoriel d'un vecteur par lui-même est zéro.

  • Le produit vectoriel est-il distributif?

    Le produit vectoriel a un effet distributif sur les vecteurs.

  • Comment détermine-t-on la direction du produit vectoriel?

    On utilise la règle de la main droite pour déterminer la direction du produit vectoriel.

  • Le produit scalaire triple doit-il être calculé d'une manière spécifique?

    Oui, il faut d'abord effectuer le produit croisé entre les deux premiers vecteurs, puis effectuer le produit scalaire avec le troisième vecteur.

  • Qu'indique un produit vectoriel nul entre deux vecteurs?

    Le produit vectoriel de deux vecteurs parallèles est nul (vecteur zéro).

  • Quel est le résultat d'un produit scalaire triple?

    Le produit scalaire triple donne un scalaire (valeur numérique) en résultat.

  • Comment trouve-t-on un vecteur unitaire perpendiculaire à un plan défini par trois points?

    Un vecteur unitaire est obtenu en divisant un vecteur par sa magnitude.

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  • 00:00:00
    we have a lesson on the cross product
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    today it's also called the vector
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    product and when we have this scenario
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    I'm going to start off with this
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    parallelogram and from this
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    parallelogram we have vector be here and
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    vector a this is also be this is also
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    vector a and this is the angle theta
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    that's made and if i consider this drop
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    this altitude if i want to find the area
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    of this rectangle I need the height
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    which from my right triangle I know this
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    is going to be the magnitude of B times
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    the sine of theta so the opposite over
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    the hypotenuse so the area of a
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    parallelogram is going to be the
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    magnitude of a time's the magnitude of B
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    sine theta and that's going to be the
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    area it is also known if you notice here
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    there is an a cross B and a cross B is
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    made to be perpendicular to B and also
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    perpendicular to a and if I go in this
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    direction it's a cross B a cross B i use
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    a right-hand rule i started a and I go
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    to be just from your physics and a cross
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    B is going to go in this direction going
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    up as I use the right hand rule the
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    magnitude of a cross B is also equal to
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    this area and so hence the air the
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    triangle is half the magnitude of a
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    cross B and that is in your formal
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    booklet but it also lends the question
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    what is this vector product idea okay
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    and so here is what I know about the
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    triangle this is in my form of accord
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    but what in the world is all this stuff
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    a cross B let's talk about that if I
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    have vector a
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    and we'll call it a 10 a to be a 1 a 2 a
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    3 and B is b1 b2 and b3 if I'm going to
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    do a cross B what I do is I set up this
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    technique and I get really good at doing
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    this going to do a lot I know I want a J
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    and K and if I tech take vector a is a1
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    a2 a3 and vector B is b1 b2 and b3 and I
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    can see from this diagram a cross B is a
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    vector so the outcome of this is going
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    to be a vector it's going to be the
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    terms of I J and K so if i want to find
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    vector I what I do here's just the
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    procedure I can use if i take this let
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    me hide this see if I can use this and
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    hide let's fill it up read and see if
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    it's hidden so if I hide this one okay
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    what I do here to find the cross product
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    is I am going to take a 2 times B 3 a 2
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    B 3 this diagonal here subtract the
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    other diagonal a3 b2 that is my I vector
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    notice I serve the top when I went down
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    when i go to my j vector the pattern
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    goes I don't start at the top I start
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    the bottom this time I want ji-hye j and
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    i'm going to go b 1 and a 3 a 3 and b 1
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    multiplied subtract the other two
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    multiplied a 1 B 3 and that is J and if
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    i want to find ki hide my k column and I
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    again I go to the top a 1 B 1 a 1 the
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    arse of a1 b2 minus a2 b1 a
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    to be one and this is the formula for
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    the cross product you will do this
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    computation a lot be very careful with
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    your signs this vector product cross
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    part is in your form liquid and you can
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    see it here this is hard for me to do
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    you'll get used to the pattern i
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    memorize a pattern well we know that
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    we're given also the magnitude is equal
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    to the sign of the theta the cross of
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    the dot product was the cosine of the
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    angle and so this will be quite a handy
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    scenario now looking at this question
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    here we want to find B cross a well if I
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    want be cross I here's I JK B cross a
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    means I have 2-4 1 and 1 32 and so I'll
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    do in combination I'll do some scratch
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    work and i'll make my B cross a vector
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    and so doing this I'm going to find I
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    first so I'm hiding that and so I end up
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    with negative four or so negative 8
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    minus 3 equals negative 11 and then I
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    move over my hiding my rectangle start
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    the pattern goes one I'll go one
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    subtract four it gives me negative 3 and
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    finally hiding k for the k i'm going to
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    go starting from the top now 6 6 minus
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    negative 4 which is 10 and so if here's
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    a and here's be the right hand rule
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    means i go in this direction and I'll a
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    cross B will be going down here this is
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    B cross a it's perpendicular here and
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    perpendicular there
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    switching colors let's do the other
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    direction now we're going to do i J and
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    K and we're going to do a cross B so
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    it's 132 and 2-4 1 i'll move my hand a
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    little rectangle to close that out and
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    so do my computations i start at the top
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    3 minus a negative 8 gives me 11 for
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    moving the rectangle over a hiding my j
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    column start at the bottom now i go 4-1
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    which gives me 3 and covering up mike a
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    column i end up with negative 4 minus 6
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    which is negative 10 and this is a cross
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    B and so if i do a cross B that means
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    I'm going in this direction here in my
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    right hand rule then take a cross B and
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    go up or this is a cross B and if I look
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    at these two vectors they are exactly
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    opposite of each other and so one thing
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    I know is that a cross B is equal to
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    minus B cross a okay so I also now want
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    a unit vector to both a and B
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    I only pause here because I think I want
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    fine the unit vector that is
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    perpendicular perpendicular I just
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    forgot that word to both a and B well if
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    I want a unit vector predictable Bambi I
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    know that both these vectors here this
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    one and this one are both perpendicular
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    to a and B and so I want to find a unit
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    vector in order to find my unit vector
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    I'm going to take this vector a here a
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    cross B and we're I'm going to get rid
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    of this that this ok so now I'm going to
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    find a unit vector so in order to do
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    that I need to find the magnitude of a
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    cross B which I know is going to be the
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    square root of 11 squared plus 3 squared
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    plus negative 10 squared which is 121
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    plus 9 plus 100 which is the square root
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    of 230 so if I want a unit vector i
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    divided by the square root of 2 30 i
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    multiply this by this value here this is
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    my unit vector now it says find the side
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    angle between a and B well I know from
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    my formula booklet this here is true you
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    pull it over this is true if I know the
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    magnitude of a cross B it will be the
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    magnitude of a time's the magnitude of B
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    sine the angle well from here a cross B
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    magnitude I figured out already is to
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    square root of 2 30
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    the magnitude of a well that's going to
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    be one plus three squared is 9 plus 2
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    squared is 4 times the magnitude of 2
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    squared plus negative 4 squared is 16
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    plus 1 squared is 1 sine of theta and so
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    if I take 2 30 / that's the square root
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    of 14 this is the square root of 21
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    equals sine of theta and so the theta
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    equals if i go to my calculator second
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    sign and i'll put it into a fraction Oh
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    try it again so second sine alpha y
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    equals four my fractions I want the
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    square root of 2 30 over square root of
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    14 and a square root of 21 and I end up
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    theta is 62.2 degrees to three
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    significant figures oh it only asked the
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    sine of the angle I did not need to go
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    this far this value would have suffice
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    for the sine of the angle let's try to
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    an example here now it says find the
  • 00:12:00
    area of parallelogram ABCD where a has
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    this coordinates well what I know about
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    parallelograms is if I they are written
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    in the order that they are so a b.a.d if
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    I want to find the area I just need this
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    vector and
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    this vector so if vector a B I do head-
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    tail so 3 minus 2 is one- two- three-
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    five- 1 minus neck up negative 1 so
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    negative 1 plus 1 is 0 ad then I go 3-2
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    is 10 minus 3 is negative 3 and then 1
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    minus a negative so one plus one is two
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    so if I want to find the area I'm going
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    to look for the cross product between
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    these and then find a magnitude so my
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    cross product JK 1 negative 50 1
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    negative 32 and so the cross product is
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    going to be here's my little rectangle
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    to help me out I want I it's going to be
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    negative 10-0 we're going to change the
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    pattern 0-2 so negative 2 hiding k start
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    of the top one so this is negative 3 and
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    negative flaps of minus 3 plus flags i
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    have to subtract minus three plus five
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    is too that is a B cross a d if i want
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    to find the area I then have to find the
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    magnitude of this magnitude of a be
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    across the magnitude of AD magnitude
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    will be 10 squared plus 2 squared plus 2
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    squared which is 104 and four so the
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    square root of 108 is the area of the
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    parallelogram find the cross product use
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    the cross product find the area alright
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    so now ABC are points find a unit vector
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    is perpendicular to the plane ABC well
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    here is a B let's say C and so there's a
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    plane here that comprises those three
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    points I want to fly a unit vector
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    perpendicular to it again I am going to
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    do the cross product of them so I'm
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    going to go a be so the head- a tail is
  • 00:15:13
    18 minus 5 is 3 9 minus 6 is 3 you
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    should pause the video and try this one
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    yourself AC and really it doesn't matter
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    which combination of vectors I choose
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    they're all on the on the plane so i can
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    choose any combination so 1 minus 2 is
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    negative 1 1 minus 5 is negative 40
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    minus 6 is minus 6 so now i'm going to
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    find a unit vector here i have to find
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    the cross product to make it
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    perpendicular so i j k 1 3 3 negative 1
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    negative 4 negative 6 now to make my
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    life easier I'm not very good with
  • 00:16:05
    negatives I make mistakes when I have
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    negatives so I'm also going to do see a
  • 00:16:11
    which is 146 it's also on the plane and
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    it will also produce a unit vector
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    that's perpendicular so will not matter
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    if it's the positive version of the
  • 00:16:22
    negative version I'm going to do the
  • 00:16:24
    positive one and so I will hide first to
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    cover up my eye and so switch to blue
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    help see it so it's 18-12 18-12 is six
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    and then next one started from the
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    bottom 3-6
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    3-6 is negative 3 and find last one I
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    start at the top 14 minus 3 4 minus 3 is
  • 00:17:01
    1 now I need a unit vector that's
  • 00:17:05
    perpendicular that's a unit vector of
  • 00:17:09
    this find the magnitude magnitude is 36
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    plus 9 plus 1 which is equal to square
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    root of 46 so if i divide by this
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    magnitude this here is a unit vector
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    that is perpendicular to this plane ABC
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    finally the properties of cross product
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    well if I take a and I cross it with a
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    because they are parallel it's
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    impossible to choose one perpendicular
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    vector so this we call it zero vector
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    and yeah it's zero vector also from
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    there if I know if a cross being equal
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    zero and a and B are parallel that's
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    comes from here I also know that a cross
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    B we figured out was equal to B cross a
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    vector also have distributive property
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    so it's a cross B plus a cross C and
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    then finally we have this last one here
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    which is called a scalar triple product
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    you have to do this first and then u dot
  • 00:18:37
    product it and it has a big formula as
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    just as easy to actually do it this is
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    called the triple scalar scalar triple
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    product scalar triple
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    product you have to do this first even
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    if it's written as such this has to be
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    done first because knees make a scammer
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    value I can't multiply a scalar times a
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    vector in this sense okay so that is
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    your cross product lots of bits the
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    important part is cross product makes a
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    vector that is parallel to both it makes
  • 00:19:20
    this angle theta and the magnitude of a
  • 00:19:23
    cross B is equal to a B sine theta where
  • 00:19:31
    this is the area of the parallelogram
Tags
  • produit vectoriel
  • parallélogramme
  • magnitude
  • angle
  • règle de la main droite
  • vecteur unitaire
  • produit scalaire triple
  • aire
  • propriétés du produit croisé
  • vecteur perpendiculaire