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Geometrical interpretation of dot product (trying to prove)

  1. Aug 14, 2009 #1
    Hi! I am trying to find out where:

    [tex]cos\theta=\frac{a \cdot b}{|a||b|}[/tex]

    came from.

    Here is mine geometrical interpretation of scalar projection:


    Now, (pr means projection)

    [tex]pr_{\overrightarrow{A}} \overrightarrow{B} = p\overrightarrow{B_0}[/tex]


    [tex]cos\theta=\frac{|pr_{\overrightarrow{A}} \overrightarrow{B}|}{|A|}[/tex]


    How do


    for 3 dimensional coordinate system, wher |a| and |b| are modulo of "a" and "b" ?

    Thanks in advance.

  2. jcsd
  3. Aug 15, 2009 #2
    Your use of the projection is recursive, since the projection of a vector use the fact that cos(theta) = dot(A,B)/|A||B|. Wikipedia offers a simple proof, look for the dot product. And the proof is valid in every dimension, not only in 2D.
  4. Aug 15, 2009 #3
    Ohh, thank you. The proof is very simple. And why my use of the projection is recursive? The scalar projection equals |A| cos(θ).
  5. Aug 15, 2009 #4
    But the dot product is hidden in |p|, and the way we compute the projection using the dot product uses the fact that we know that it's equal to cos(theta)/|A||B|. Explicitly, it looks like this:

    1- Hypothesis: A dot B = cos(theta)/|A||B|
    2- Use it to define projection
    3- Use projection arguments to prove 1.
  6. Aug 15, 2009 #5


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    An inner product space is a particular type of normed spaces, thus it makes little meaning to say that the inner product is "hidden" within the concept of length.
  7. Aug 15, 2009 #6
    Here, p is a scalar, that comes magically from the projection. And if it don't use the dot product to compute p, then I don't see the link between the dot product and cos(theta).
  8. Aug 15, 2009 #7
    Thanks for the replies. Now I understand the dot product. But what about the cross product?


    [tex]\overrightarrow{c} \cdot \overrightarrow{a} = \overrightarrow{0}[/tex]


    [tex]\overrightarrow{c} \cdot \overrightarrow{b} = \overrightarrow{0}[/tex]


    How is possible that the area of the parallelogram is the magnitude of a x b or c ??

  9. Aug 15, 2009 #8


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    The length of a cross product, [itex]a\times b[/itex] is, as you say, [itex]|a||b|sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between vectors a and b. In addition, the area of a parallelogram is "height times base" where "height" is measured perpendicular to the "base". If you take, for example, the vector b as base, then you can drop a perpendicular from the tip of a to b to get a right triangle- the length of the perpendicular is "height" and, since the right triangle has angle [itex]\theta[/itex] and hypotenuse of length |a|, that perpendicular is the "opposite leg" so [itex]h/|a|= sin(\theta)[/itex] and [itex]h= |a|sin(\theta)[/itex].
  10. Aug 16, 2009 #9
    Thanks for the reply.
    Yes, I know that the area of parallelogram is A=aha, and [itex]sin\theta=\frac{h_a}{b}[/itex], so that [itex]h_a=bsin\theta[/itex] and [itex]A=absin\theta[/itex]. What I do not understand is why the vector "c" got magnitude of the area of the parallelogram between a and b, i.e the magnitude of the vector is [itex]A=absin\theta[/itex]. Why the magnitude isn't a*b or some other value?

  11. Aug 16, 2009 #10
    Using Lagrange's identity identity, you get

    [tex]|a \times b|^2 = (a \cdot a)(b \cdot b) - (a \cdot b)(a \cdot b)[/tex]

    Which is simply equal to

    [tex]|a \times b|^2 = |a|^2|b|^2 - |a|^2|b|^2cos^2(\theta)[/tex]
    [tex]|a \times b|^2 = |a|^2|b|^2 (1 - cos^2(\theta))[/tex]
    [tex]|a \times b|^2 = |a|^2|b|^2 sin^2(\theta)[/tex]
  12. Aug 16, 2009 #11
    Thanks for the reply. I know that, but why the magnitude of the a x b, is [itex]|a||b|sin\theta[/itex]? If we suppose that c is vector so that it is normal to a and b (i.e c . a =0 and c.b=0) and a x b = c, why the magnitude of c (i.e |c| ) is [itex]|c|=|a||b|sin\theta[/itex]??

  13. Aug 16, 2009 #12
    Then I really don't see what you want to know. I proved by an algebraic property that [tex]|c|=|a||b|sin\theta[/tex] and I did not suppose anything.
  14. Aug 16, 2009 #13
    Yes, its actually that's what I do not understand. Look at the picture. The magnitude of |c| equals the area of the parallelogram.

    I already proved that [tex]A=|a||b|sin\theta[/tex] where A is the area of the parallelogram. But what I do not understand how A and c are linked together?


    P.S Here is thehttp://upload.wikimedia.org/wikipedia/commons/thumb/4/4e/Cross_product_parallelogram.svg/180px-Cross_product_parallelogram.svg.png" [Broken]
    Last edited by a moderator: May 4, 2017
  15. Aug 16, 2009 #14
    ok! So we need to do it in three steps:

    1- Show that the area of a parallelogram in 2D is the determinant of a 2x2 matrix containing the two side vectors.
    2- Show that the area of a parallelogram in 3D is the squared sum of his projection on the three axis planes.
    3- Make the final link.

    1- I won't do it here, because it's long, but you can look here: http://mathforum.org/library/drmath/view/71204.html . Geometric/Algebraic proof, the boring part.

    2- We now project the parallelogram on the 3 axis planes (YZ-plane, ZX-plane, XY-plane) by eliminating the components corresponding to the projection axis. E.g. if [itex]A=(a_x,a_y,a_z), B=(b_x,b_y,b_z)[/itex], then [itex]A_x=(0,a_y,a_z), B_x=(0,b_y,b_z)[/itex]. The area of a projected parallelogram is simply the determinant of the 2x2 matrix containing the remaining components. You need to follow the even permutation rule to put the components in the matrix. E.g. For X, put Y first, then Z. For Y, put Z then X. For Z, put X then Y. For what follows, P is the area of the 3D parallelogram, and [itex]P_x,P_y,P_z[/itex] are the areas of the projected parallelograms.

    Here's the most interesting part. In 3 dimensions, there's a duality between a line and a plane (This duality can be generalized in every dimension but let's concentrate on this particular case, for more, look at Exterior Algebra). And we can compute the area of a parallelogram with a Pythagorean-like formula:


    Without going into endless algebraic computation, you can convince yourself that it's correct by looking at the trivial cases where the parallelogram is parallel to a axis plane. Then consider a rectangular shape inclined in only one direction. Once you pictured this second example, the formula should be clear.

    3- If we look carefully, [itex]P=(P_x,P_y,P_z)=A \times B[/itex], so each component of the cross product contains the area of the parallelogram projected

    I don't think I can go more geometric than this.
  16. Aug 18, 2009 #15
    Sorry for being late, but I got problems on college.

    1. Yes, I completely understand it.

    2. Ohhh, I know what you mean.

    [tex]\overrightarrow{a} \times \overrightarrow{b} = (\begin{vmatrix}
    a_y & a_z \\
    b_y & b_z
    \end{vmatrix}, \begin{vmatrix}
    a_z & a_x \\
    b_z & b_x
    \end{vmatrix}, \begin{vmatrix}
    a_x & a_y \\
    b_x & b_y
    \end{vmatrix} )[/tex]

    I know how to prove it by solving:

    [tex]\overrightarrow{a} \times \overrightarrow{b}=(a_x \overrightarrow{i} +a_y \overrightarrow{j} + a_z \overrightarrow{k}) \times (b_x \overrightarrow{i} +b_y \overrightarrow{j} + b_z \overrightarrow{k})[/tex]

    Thank you very much.

    Anyway, do you have some picture for the projection of the parallelogram on the planes?

    Can somebody explain why:

    [tex]\overrightarrow{i} \times \overrightarrow{k} = - \overrightarrow{j}[/tex]

    and not

    [tex]\overrightarrow{i} \times \overrightarrow{k} = \overrightarrow{j}[/tex]

    Is this taken by convention or something?

  17. Aug 18, 2009 #16
    Sorry, I don't have any picture, but it's simple to draw it on paper. Just take the YZ-plane projection and draw the two vectors [itex](a_y,a_z), (b_y,b_z)[/itex]. It's like when you look at it from the X axis.

    [itex]\overrightarrow{i} \times \overrightarrow{k} = - \overrightarrow{j}[/itex] because it's the algebraic area, which can be negative. This has to do with the orientation of the parallelogram which has two sides. It's not a convention, since you can compute it with [itex]\overrightarrow{i} \times \overrightarrow{k} = (1,0,0) \times (0,0,1) = (0,-1,0)[/itex]. This come from the definition of the cross product.
  18. Aug 19, 2009 #17
    Thanks, I know that is simple to draw it on paper, but I can't flag the answer [itex]P=\sqrt{P_{x}^{2} + P_{y}^{2}+P_{z}^{2}}[/itex] as obvious.

    Here is the picture:


    I chose random vectors a = (2,1,-1) and b = (-1,2,1).

    The green one is |a x b|, the other ones are the parallelograms in XY, XZ and YZ planes.

    I made |i|=|j|=|k|=1 cm (you can see on the diagram small dots on the x,y,z axis)

    Now for the second question.

    Also I understand the cross product respect to the right-handed coordinate system:

    http://upload.wikimedia.org/wikipedia/commons/thumb/b/b0/Cross_product_vector.svg/180px-Cross_product_vector.png [Broken]

    i = (1,0,0) and k=(0,0,1)

    So c. i = (a,b,c).(1,0,0)=0

    which a=0


    which gave us c=0

    Now we got the normal vector c=(0,b,0)

    Since we know that |c|=1 we can conclude that |b|=1, which [itex]b=\pm 1[/itex]

    Now we need to get in mind the (iii) rule, and that is the right-handed coordinate system.

    Here I am stuck.

    Last edited by a moderator: May 4, 2017
  19. Aug 19, 2009 #18
    Take an easier example, like the rectangular inclined parallelogram define by the two vectors [itex]\vec{A}=(a,0,0), \vec{B}=(0,b,c)[/itex]. Only two projections has non-null area. Defining [itex]h=\sqrt{b^2 + c^2}[/itex], we know that [itex]P = ah[/itex] since it's rectangular. But

    P = ah = a\sqrt{b^2 + c^2} = \sqrt{(ab)^2 + (ac)^2} = \sqrt{P_z^2 + P_y^2}

    That's is really not a proof by I think it's enough to be convinced of the geometry behind this formula.

    You don't have to use you fingers to know that it's equal to -1, just do some matrix determinants.

    (1,0,0)\times(0,0,1) = \left( \begin{vmatrix}0&0\\0&1\end{vmatrix}, \begin{vmatrix}0&1\\1&0\end{vmatrix}, \begin{vmatrix}1&0\\0&0\end{vmatrix} \right) = (0,-1,0)
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