Dot product scalar distributivity

In summary, the formula ##(\alpha\vec{a})\cdot b = \alpha(\vec{a}\cdot\vec{b})## is correct and does not need to be changed to ##|\alpha|(\vec{a}\cdot\vec{b})##. This is because the angle between vectors changes when using a negative scalar value for ##\alpha##.
  • #1
archaic
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I'm having a little trouble with this :
We have ##(\alpha\vec{a})\cdot b = \alpha(\vec{a}\cdot\vec{b})## but shouldn't it be ##|\alpha|(\vec{a}\cdot\vec{b})## instead since ##||\alpha \vec{a}||=|\alpha|.||\vec{a}||## ?

##(\alpha\vec{a})\cdot b = ||\alpha\vec{a}||.||\vec{b}||.\cos\theta = |\alpha|.||\vec{a}||.||\vec{b}||.\cos\theta = |\alpha|(\vec{a}\cdot\vec{b})##
 
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  • #2
archaic said:
I'm having a little trouble with this :
We have ##(\alpha\vec{a})\cdot b = \alpha(\vec{a}\cdot\vec{b})## but shouldn't it be ##|\alpha|(\vec{a}\cdot\vec{b})## instead since ##||\alpha \vec{a}||=|\alpha|.||\vec{a}||## ?

##(\alpha\vec{a})\cdot b = ||\alpha\vec{a}||.||\vec{b}||.\cos\theta = |\alpha|.||\vec{a}||.||\vec{b}||.\cos\theta = |\alpha|(\vec{a}\cdot\vec{b})##

Clearly ##(-\vec{a})\cdot b = -(\vec{a}\cdot\vec{b})##

Where you have gone wrong is assuming that ##\theta## is the same angle between vectors ##\vec{a}, \vec{b}## and ##\alpha \vec{a}, \vec{b}##. If you draw a diagram for 2D vectors with ##\alpha = -1## you'll see that this changes the angle.
 
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  • #3
PeroK said:
Clearly ##(-\vec{a})\cdot b = -(\vec{a}\cdot\vec{b})##

Where you have gone wrong is assuming that ##\theta## is the same angle between vectors ##\vec{a}, \vec{b}## and ##\alpha \vec{a}, \vec{b}##. If you draw a diagram for 2D vectors with ##\alpha = -1## you'll see that this changes the angle.
Right, thank you!
 

What is the dot product scalar distributivity property?

The dot product scalar distributivity property is a mathematical property that states that the dot product of a scalar and a vector is equal to the scalar multiplied by each component of the vector. This can be represented as: c * (a, b, c) = (c*a, c*b, c*c).

How is the dot product scalar distributivity property used in vector algebra?

The dot product scalar distributivity property is used in vector algebra to simplify calculations involving scalars and vectors. It allows us to distribute the scalar across each component of the vector, making it easier to manipulate and solve equations involving both scalars and vectors.

What is the significance of the dot product scalar distributivity property in physics?

In physics, the dot product scalar distributivity property is used to calculate work done on an object by a force. The dot product of the force vector and the displacement vector is equal to the work done, and the distributivity property allows us to simplify this calculation by distributing the scalar representing the force across each component of the displacement vector.

Can the dot product scalar distributivity property be extended to higher dimensions?

Yes, the dot product scalar distributivity property can be extended to higher dimensions. In three-dimensional space, for example, the dot product of a scalar and a vector would be equal to the scalar multiplied by each component of the vector, just like in two dimensions. This can be represented as: c * (a, b, c) = (c*a, c*b, c*c).

How does the dot product scalar distributivity property relate to the commutative and associative properties?

The dot product scalar distributivity property is related to the commutative and associative properties in that it follows the same rules. The order in which we perform scalar multiplication does not matter, and we can also distribute a scalar across a sum of vectors. This allows us to manipulate equations and terms in a similar way to how we would with scalars using the commutative and associative properties.

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