# Dot product scalar distributivity

• B
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})##

## Answers and Replies

Homework Helper
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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})##

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
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!