Matematické Fórum

stuart clark · 11. 02. 2015 04:24

Evaluation of Three- Dimensional vectors $\vec{v_{1}\;}\;,\vec{v_{2}}\;,\vec{v_{3}}$ satisfying $\vec{v_{1}}\cdot \vec{v_{1}} = 4\;,\vec{v_{1}}\cdot \vec{v_{2}} = -2\;,\vec{v_{1}}\cdot \vec{v_{3}}=6$

and $\vec{v_{2}}\cdot \vec{v_{2}} = 2\;,\vec{v_{2}}\cdot \vec{v_{3}} = -5\;,\vec{v_{3}}\cdot \vec{v_{3}}=29$ is

check_drummer · 12. 02. 2015 00:50

Hi,
I will only start: wlog v1:=(2,0,0) adding first and 2x second equality gives v1.(v1+2v2)=0 ie v1+2v2 is perpendicular to v1, ie y and z component of v1+2v2 is 0 so wlog v2=(x,y,0) and values x,y (there are 2 possibilities) can be easily found from previous and from the equation of the norm of v2. In the similar way we obtain v3 in two ways - so for given v1 we get 4 possibilities (to get all v1,v2,v3) - and all possibilities will be obtained choosing direction of v1 arbitrarily.

xxMari · 12. 02. 2015 22:54

Autor skryl část textu. Zobrazit/skrýt!

stuart clark · 19. 02. 2015 17:56

Thanks ↑ xxMari:↑ check_drummer:

To ↑ check_drummer: I did not understand the solution, Why we take $v_{1} = (2,0,0)$ Thanks

To ↑ xxMari: , Would you like to explain me the Gram matrix. Thanks

check_drummer · 19. 02. 2015 22:20

↑ stuart clark:
Hi, because the norm of v1 is 2 (v1.v1=4).

stuart clark · 23. 02. 2015 14:31

Thanks ↑ check_drummer: got it.

Matematické Fórum

#1 11. 02. 2015 04:24

vectors

#2 12. 02. 2015 00:50

Re: vectors

#3 12. 02. 2015 22:54

Re: vectors

#4 19. 02. 2015 17:56

Re: vectors

#5 19. 02. 2015 22:20

Re: vectors

#6 23. 02. 2015 14:31

Re: vectors

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