Confused about dot product of a and b = |a||b| if theta = 0

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SUMMARY

The discussion centers on the confusion surrounding the dot product of two vectors, specifically when the angle θ between them is 0. The dot product is defined as a·b = |a||b|cos(θ), which simplifies to a·b = |a||b| when θ = 0, indicating that the vectors are parallel. However, participants pointed out that the original calculations presented by the user contained typos and misunderstandings regarding the definitions of the dot product and the magnitudes of the vectors. The correct interpretation confirms that when θ = 0, the dot product indeed equals the product of the magnitudes of the vectors.

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annamal
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I am not sure what I am doing wrong but dot product of a and b =/= |a||b| when I am trying to calculate it
I am not sure what I am doing wrong but dot product of a and b =/= |a||b| when I am trying to calculate it. Theta = 0:
dot product(a and b) = ax*bx + ay*by
|a||b|= sqrt((ax^2+ay^2)*(ax^2 + by^2)) = sqrt((ax*bx)^2 + (ax*by)^2 + (ay*bx)^2 + (ay*by)^2) =/= ax*bx + ay*by

What am I doing wrong?
 
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These aren't supposed to be equal.
$$a\cdot b = ||a|| ||b|| \cos(\theta)$$
Where ##\theta## is the angle between ##a## and ##b##.
 
Office_Shredder said:
These aren't supposed to be equal.
$$a\cdot b = ||a|| ||b|| \cos(\theta)$$
Where ##\theta## is the angle between ##a## and ##b##.
I'm not sure I get what you're saying.
a dot b = ||a|| ||b|| with theta = 0
 
annamal said:
Summary:: I am not sure what I am doing wrong but dot product of a and b =/= |a||b| when I am trying to calculate it

I am not sure what I am doing wrong but dot product of a and b =/= |a||b| when I am trying to calculate it. Theta = 0:
dot product(a and b) = ax*bx + ay*by
|a||b|= sqrt((ax^2+ay^2)*(ax^2 + by^2)) = sqrt((ax*bx)^2 + (ax*by)^2 + (ay*bx)^2 + (ay*by)^2) =/= ax*bx + ay*by
If theta = 0, a and b are in the same direction, so ax/ay = bx/by.
 
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The theta referred to in the dot product definition is the angle between the vectors when placed next to each other with the tails touching.

As mentioned in earlier posts, you can’t just pick two vectors and declare that theta is zero unless you know the two vectors are parallel ie point in the same direction.
 
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annamal said:
Summary:: I am not sure what I am doing wrong but dot product of a and b =/= |a||b| when I am trying to calculate it

dot product(a and b) = ax*bx + ay*by
There are two definitions for the dot product, assuming ##\vec a = <a_1, a_2>## and ##\vec b = <b_1, b_2>## .
Coordinate definition: ##\vec a \cdot \vec b = a_1b_1 + a_2b_2##
Coordinate-free definition: ##\vec a \cdot \vec b = |\vec a||\vec b|\cos(\theta)##, where ##\theta## is the angle between the two vectors, and ##|\vec a|, |\vec b|## are the magnitudes of the two vectors.
 
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Office_Shredder said:
These aren't supposed to be equal.
$$a\cdot b = ||a|| ||b|| \cos(\theta)$$
Where ##\theta## is the angle between ##a## and ##b##.
Did you mean "are supposed to be equal"? The equation above is one of the definitions of the dot product.
 
FactChecker said:
It is not a simple proof, even when the vectors are aligned. You can compare your steps to the proof in Case 2 here: https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product
On second thought, I think it should be simple. Since the theta is 0, one is a constant multiple of the other. Let ##b = ca##. Then
##a\cdot ca = a_xca_x+a_yca_y = c(a_x^2+a_y^2)##.
And ##|a||ca| = c|a|^2 = c(a_x^2+a_y^2)##.
 
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  • #11
From post #1:
annamal said:
dot product(a and b) = ax*bx + ay*by
|a||b|= sqrt((ax^2+ay^2)*(ax^2 + by^2)) = sqrt((ax*bx)^2 + (ax*by)^2 + (ay*bx)^2 + (ay*by)^2) =/= ax*bx + ay*by

What am I doing wrong?
For one thing, there are typos in your equation that I don't think others have noticed. Also, it's unclear to me how your vectors are defined. I'm going to guess that ##\vec a = <a_x, a_y>## and ##\vec b = <b_x, b_y>##. If that's the case, then ##| \vec a| = \sqrt{a_x^2 + a_y^2}## and ##| \vec b| = \sqrt{b_x^2 + b_y^2}##.

Your expression |a||b|= sqrt((ax^2+ay^2)*(ax^2 + by^2)) has a typo. The second factor in your square root should be ##b_x^2 + b_y^2##, not ##a_x^2 + b_y^2##. This typo appears again in the equation I quoted.

mcastillo356 said:
I think your doubts come when unit vectors take place, and I think you are not asking about dot product, but vector product in components
I don't think either of the above is true. There's no indication that the vectors involve are unit vectors, nor is there any indication that I see that the vector product (AKA cross product) is intended, especially since the vectors are two-dimensional.
 
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  • #12
I figured it out guys.
|a||b|= sqrt((ax^2+ay^2)*(bx^2 + by^2)) = sqrt((ax*bx)^2 + (ax*by)^2 + (ay*bx)^2 + (ay*by)^2) = ax*bx + ay*by since axaxbyby + ayaybxbx = axaxbyby + (axby/bx)^2*bx^2 = axaxbyby + (axby)^2 = 2axaxbyby due to ay/ax = by/bx
Thanks though.
 
  • #13
If ##\theta =0##, then ##\vec b = \alpha \vec a##, for some ##\alpha > 0##. And$$\vec a \cdot \vec b = \alpha(\vec a \cdot \vec a) =\alpha|\vec a|^2 = |\vec a||\alpha \vec a| = |\vec a||\vec b| $$Where$$\vec a \cdot \vec a = a_x^2 + a_y^2 = |\vec a|^2$$
 
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  • #14
FactChecker said:
It is not a simple proof, even when the vectors are aligned. You can compare your steps to the proof in Case 2 here: https://proofwiki.org/wiki/Cosine_Formula_for_Dot_Product
The second case is poorly written. At a quick glance, case 2 looks like a complete proof, which it is not. At the top of the page it says that it is for vectors that are scalar multiples of each other. Within the case, it says that without loss of generality that they are scalar multiples of each other. It also uses that the vectors are scalar multiples of each other to equate sgn(c) and cosθ by implication.
 
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  • #15
caz said:
The second case is poorly written. At a quick glance, case 2 looks like a complete proof, which it is not. At the top of the page it says that it is for vectors that are scalar multiples of each other. Within the case, it says that without loss of generality that they are scalar multiples of each other. It also uses that the vectors are scalar multiples of each other to equate sgn(c) and cosθ.
The OP stipulated that the vectors were at an angle ##\theta=0##. So they are scalar multiples of each other.
 
  • #16
FactChecker said:
The OP stipulated that the vectors were at an angle ##\theta=0##. So they are scalar multiples of each other.
I am talking about the website proof. It is poorly written.
 
  • #17
caz said:
The second case is poorly written. At a quick glance, case 2 looks like a complete proof, which it is not. At the top of the page it says that it is for vectors that are scalar multiples of each other. Within the case, it says that without loss of generality that they are scalar multiples of each other. It also uses that the vectors are scalar multiples of each other to equate sgn(c) and cosθ by implication.
It doesn't say "Proof 2", it says "Case 2", the second case is that the vectors are scalar multiples of each other. So I read it as: Given that the vectors are scalar multiples of each other, wolog let ##v = cw##, where ##c## is some scalar. I thought that was sufficiently well-stated.

EDIT: I guess a good practice would have been to restate what the conditions of Case 2 were at the beginning of Case 2. (Maybe as a sort of header.)
 

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