How can I find the angle between the 3d vector & x-axis ?

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    3d Angle Vector
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To find the angle between the vector 3i + j - 4k and the x-axis, the dot product method is used. The angle can be calculated using the formula cos(θ) = (A · B) / (|A| |B|), where A is the vector and B is the unit vector along the x-axis. The dot product of the two vectors is computed, and the magnitudes of the vectors are also determined. The discussion highlights the importance of understanding the dot product concept for solving such problems. Overall, the method provides a clear approach to finding angles between vectors in 3D space.
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Hello :- My name is Anas I'm a new member in this forum .

I have a small question

what is the angle between the vector 3i +j-4k and the x-axis ?

and please explain me what is the method of answer this question

Im sorry , I know I have many mistakes in this topic because my English language is poor

Thank you , and I wish from you to answer me very fast
 
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Tell us what you know about the Dot product of 2 vectors.
 
I Didnt know about this
Im sorry
 
Maybe you know it as the http://en.wikipedia.org/wiki/Dot_product"
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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