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

[annamal]

TL;DR 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

What am I doing wrong?

 

[Office_Shredder]

These aren't supposed to be equal.
$$a\cdot b = ||a|| ||b|| \cos(\theta)$$
Where $\theta$ is the angle between $a$ and $b$.

 

[annamal]

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

 

[willem2]

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.

 

[jedishrfu]

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.

 

[Mark44]

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.

 

[Mark44]

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]

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

 

[FactChecker]

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 $|a||ca| = c|a|^2 = c(a_x^2+a_y^2)$.

 

[mcastillo356]

@annamal , here is some bakground on vectors. 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
Hint: Solution to yor exercise: zero
https://users.physics.ox.ac.uk/~harnew/lectures/lecture2-handout.pdf
Love

 

[Mark44]

From post #1:

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.

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.

 

[annamal]

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.

 

[PeroK]

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$$\vec a \cdot \vec a = a_x^2 + a_y^2 = |\vec a|^2$$

 

[Frabjous]

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.

 

[FactChecker]

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.

 

[Frabjous]

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.

 

[FactChecker]

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.)