# Dot product scalar distributivity

• B

## Main Question or Discussion Point

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
Homework Helper
Gold Member
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!