B Dot product scalar distributivity

[archaic]

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 $||\alpha \vec{a}||=|\alpha|.||\vec{a}||$ ?

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

 

[PeroK]

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.

 

[archaic]

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!