December 28th, 2018, 10:08 PM  #1 
Newbie Joined: Aug 2018 From: România Posts: 22 Thanks: 2  A vectorial calculation
Hello all, Knowing $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ to calculate $\displaystyle \overrightarrow{z}^n$ where the three vectors are not collinear , $\displaystyle n\in \mathbb N$ and $\displaystyle n\geq 2 $. All the best, Integrator Last edited by Integrator; December 28th, 2018 at 10:29 PM. 
December 28th, 2018, 10:45 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
what does $\large \vec{z}^n$ mean?

December 29th, 2018, 02:26 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,663 Thanks: 649 
What are you given?

December 29th, 2018, 03:05 PM  #4 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 
The title of the thread is ”vectorial calculation” Maybe n is an exponent 
December 29th, 2018, 04:24 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  
December 29th, 2018, 07:11 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198  
December 29th, 2018, 09:51 PM  #7 
Newbie Joined: Aug 2018 From: România Posts: 22 Thanks: 2  Heloo, This is a trivial case....The scalar product $\displaystyle \overrightarrow {v}^n=\overrightarrow{v} \cdot \overrightarrow{v} \cdots \overrightarrow{v}\cdot \overrightarrow{v}$ is what would be interesting to calculate. All the best, Integrator 
December 29th, 2018, 11:00 PM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198  Quote:
$\vec{v}\cdot \vec{v}$ is a scalar and thus cannot be further dotted with a vector. Please stop wasting everyone's time.  
December 30th, 2018, 03:58 AM  #9 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235 
A relevant product would be the geometric algebra product, which for vectors equals $$\overrightarrow{v}\overrightarrow{w} = \overrightarrow{v}\cdot \overrightarrow{w} + \overrightarrow{v}\times \overrightarrow{w},$$ the sum of cross and dot product. But to compute further and take the $n$th power of a vector, you'd need the formula for the geometric algebra product for arbitrary blades, and that's pretty ugly. 
December 30th, 2018, 10:47 PM  #10  
Newbie Joined: Aug 2018 From: România Posts: 22 Thanks: 2  Quote:
The problem I posted is from another forum...It is known that if $\displaystyle \overrightarrow{z}=\overrightarrow{x}+ \overrightarrow{y}$ , then $\displaystyle \overrightarrow{z} ^2=\overrightarrow{x}^2+\overrightarrow{y}^2+2 \overrightarrow{x}\overrightarrow{y}$ and so what expressions do they have $\displaystyle \overrightarrow{z} ^3$ , $\displaystyle \overrightarrow{z}^4$ , $\displaystyle \overrightarrow{z}^5$ ,......., $\displaystyle \overrightarrow{z}^{n1}$ , $\displaystyle \overrightarrow{z}^n$? Thank you very much! All the best! Integrator  

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