Finding the Angle Between Two Vectors A & B

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To find the angle between vectors A = -2i + 5j and B = 2i + 3j, the cross product A x B can be calculated, which results in a vector. The formula for the angle between two vectors involves the sine of the angle, expressed as |A x B| = |A||B|sin(θ). Confusion arose regarding the calculation of the cross product and its components, specifically with the negative signs in the determinant setup. Understanding the relationship between the vector product and the angle will clarify the solution process.
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Homework Statement



Two vectors are given by A = -2 i + 5 j and B = 2 i + 3 j
Also find the angle between them

Homework Equations



Not really sure but from my book i get A x B = -B x A

The Attempt at a Solution



the answer is -16, but I'm confused I thought that it should look like -6 + 10 but why is it minus ten?

How would I use this to find the angle?
 
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Do you know how to set up a determinant for vector product? From the determinant you can easilly calculate the product vector. Remember the result of a vector product is always a vector.

When you know the vector product you can find the angle from

(AxB)=|A||B|sin(A,B)

Where A and B are the 2 vectors, (AxB) is the vector product and |A|/|B| are the lenghts of the vectors.
 
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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