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Dot product

  • Thread starter mbrmbrg
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  • #1
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Nice, simple, straighforward problem:
[itex]j \cdot {-j}[/itex]

Difficulty: I solved it two ways and got two answers.

[itex]\vec{a} \cdot \vec{b}=ab\cos\phi[/itex]

So in this case, I should get [tex]j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex]

However, using the formula [itex]\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}[/itex], I got [itex]j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1[/itex]

Can anyone tell me why?
 
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Answers and Replies

  • #2
809
0
mbrmbrg said:
Nice, simple, straighforward problem:
[itex]j \cdot {-j}[/itex]

Difficulty: I solved it two ways and got two answers.

[itex]\vec{a} \cdot \vec{b}=ab\cos\phi[/itex]

So in this case, I should get [tex]j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex]

However, using the formula [itex]\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}[/itex], I got [itex]j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1[/itex]

Can anyone tell me why?
Recall that the dot product is:
[tex] \vec A \cdot \vec B = |\vec A| |\vec B| \cos \theta_{AB} [/tex]

let [itex] \vec A = (0,-1,0) [/itex]
[tex] A=|\vec A| = \sqrt{(0)^2 +(-1)^2 +(0)^2} [/tex]

What is [itex] (-1)^2 [/itex] :)
 
  • #3
486
1
Neither my book nor my professor mentions the absolute value thing. I guess that's completely implied, though, because the negative sign has to do with direction, not magnitude.

Thank you so much!
 
  • #4
Astronuc
Staff Emeritus
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[tex] |\vec A|[/tex] means magnitude of [tex] \vec A[/tex], which is a scalar and is given by the square root of the sum of the squares of the magnitudes of the component vectors. [tex]|\vec{a}|\,=\,\sqrt{a^2_x+a^2_y+a^2_z}[/tex]



One could also use the fact that -j * j = - (j * j) = - (1) = -1, where I use * to mean the dot product.

Presumable j is a unit vector (0, 1, 0).

Back to the OP:

[tex]j \cdot {-j}=(1)(1)(cos(180^o))=(1)(1)(-1)=-1[/tex]
 
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  • #5
I think basically your mistake lies in your calculation of the modulus(magnitude) of -j which is 1 not -1.For use in the formula a*b=abcos#
where * means dot product
a/b mean the magnitudes(moduli) of a & b
 
  • #6
486
1
Astronuc said:
One could also use the fact that -j * j = - (j * j)
How is that derived/proven?
 
  • #7
809
0
mbrmbrg said:
Astronuc said:
One could also use the fact that -j * j = - (j * j)

How is that derived/proven?
The dot product operation has an associative property, that says:

[tex] (k \vec A) \cdot \vec B = k (\vec A \cdot \vec B) [/tex]
Where [itex] k [/itex] is a scalar, and [itex] \vec A, \,\, \vec B [/itex] are vectors.
 
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  • #8
486
1
Riiiiiight....
I've got to start consistently thinking of a negative sign as -1 multiplied by what follows, don't I.
Thanks!
 
  • #9
809
0
mbrmbrg said:
Riiiiiight....
I've got to start consistently thinking of a negative sign as -1 multiplied by what follows, don't I.
Thanks!
I remember how much grief that used to give me :smile:
 

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