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Find a vector perpendicular to Oy axis. Knowing that v*v1=8 and v*v2=-3. v1=(3,1,-2) and v2=(-1,1,1)

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A bit confused by this question, is this what you have written above the dot product? the x-axis is perpendicular to the y-axis, also the z-axis
...or any other vector in the without a y component
oh wait, it's supposed to be perpendicular to which axis? @viniterranova please explain so we can help you

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Other answers:

Determine the vector v, orthogonal to the axis Oy.
where v*v1=8 and v*v2=-3
you did not really clarify you just repeated yourself are these dot products?
Yes these are dot products
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then the y component can be left away, will result in a system of equations.
Here it´s a shot from the book where i taken the question.
so we have\[\vec v\cdot\vec v_1=8\]\[\vec v\cdot\vec v_2=-3\]where\[\vec v_1=\langle3,1,-2\rangle\]\[\vec v_2=\langle-1,1,1\rangle\]and if it is perpendicular to the y axis we have that\[\vec v\cdot\langle0,1,0\rangle=0\]let\[\vec v=\langle x,y,z\rangle\] and we get the system of equations spacelimbus was referring to
And, what´s more?
\[\vec v\cdot\vec v_1=3x+y-2z=8\]\[\vec v\cdot\vec v_2=-x+y+z=-3\]\[\vec v\cdot\langle0,1,0\rangle=y=0\]
So, in this case y = zero? That ´s right? So, i just plug 0 in the equation in order to slove the system? Am i right?
In this i got a following system 3x-2z=8 and -x+z= -3 and solving in function of z i got z=x-3 and so, i have reached x=2. Am i right?
x=2, z=-1, is what I got if you want to compare.
Yes, that´s for sure. x=2 and z=-1. Thanks guys.
So the answer to the question is w=(2,0,-1)
well in the exercise they called the vector v, so you maybe want to name him like that too, but that's not a convention, the coordinates are what's important, and you maybe want to write it in brackets, see how @TuringTest wrote the vectors.
Thanks for the help. But, i don´t know how to use the Latex yet. Thanks a lot for the help.

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