Dot product scalar distributivity

  • #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
PeroK
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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.
 
  • #3
archaic
688
210
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!
 

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