What is the Dot Product of Unit Vectors in Vector Projection?

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The discussion focuses on understanding the dot product of unit vectors in the context of vector projection. The example provided involves two unit vectors, u' and v', derived from a force vector of 7 kN. The confusion arises from why the dot product of these unit vectors is multiplied by the force's magnitude. It is clarified that this process helps find the component of the force in the direction of the unit vector v'. The final takeaway is that the dot product is essential for determining how much of the force aligns with the direction of the vector being considered.
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I was copying my friends notes and had a hard time understanding one of the examples he had written down from lecture. See the attachment for a the picture of the example. This example looks like a projection of two vectors to me, but I'm not sure.

u'=\frac{4i+2j}{\sqrt{20}} u' = unit vector u in the direction of force
v'=\frac{3i+4j}{5} v' = unit vector v

Fy&#039; = 7kN (\frac{4i+2j}{\sqrt{20}})\cdot (\frac{3i+4j}{5}) <br /> <br /> = 7kN (\frac{4}{\sqrt{20}} * \frac{3}{5} + \frac{2}{\sqrt{20}} * \frac{4}{5})

The unit vectors came from the drawing and are in the direction of the two vectors. The thing i don't get is why are 2 unit vectors being dotted, then multiplied by the magnitude 7 kN?

http://img394.imageshack.us/img394/5998/untitled3bz.png

**EDIT: forgot to include in the image a unit vector symbol in the pic for F, so F = 7 kN e' (e' is a unit vector)
 
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It is the same regardless of how you do it. If you multiply out the magnitude of the F vector onto the F's unit vector, you get the original vector F. You can then dot it against the y vector to find F \cdot y [/tex] (which I think is what your question is asking).
 
The subscript on the Fy should be v' instead! (ie, your e' = v')
that is, your notes found the v' component of the 7kN Force
which original vector was along the u' direction.

F_y would be = F cos(phi), where cos(phi) = u'.j (<=dot product)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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