# [[another]] question about projections

1. Sep 2, 2008

### imsoconfused

OK, this question has a lot of parts, so I'll do my best to write it in a way that makes sense. Basically, aVR, AVL, and aVF are the three main vectors in the direction of blood being pumped by the heart. The three endpoints of the vectors form an equilateral triangle that goes from each hand to the feet and hand to hand. These are the leads labeled LeadI, LeadII, and LeadIII. Looking at the body from the front, LeadI stretches from the left side (right hand) to the right side (left hand), LeadII stretches from the left side to the bottom (the feet) and LeadIII stretches from the right side (left hand) to the bottom (the feet). aVR lead is -(1/2)LeadI -(1/2)LeadII, (aVL=(1/2)LeadI - (1/2)LeadIII, and aVF=(1/2)LeadII+(1/2)LeadIII, FYI). The total heart vector is V=2i-j, and we need to find the projeciton on the aVR lead of V. Lead I is the horizontal vector 2i, so V.LeadI=4. Lead II is the -60deg vector i - sqrt(3)j, so V.LeadII=2+sqrt(3). Lead III is the -120deg vector -i - sqrt(3)j, so V.LeadIII= -2 + sqrt(3).

If that wasn't confusing enough, here is what I have used/done.

P=((A.B)/|A|^2)A so here it would be P=((V.aVR)/|V|^2)V. aVR= -(1/2)LeadI - (1/2)LeadII and V=2i-j. That means:

V.(-(1/2)LeadI-(1/2)LeadII = (V.(-1/2)LeadI) + (V.(-1/2)LeadII) = 4 + 2 + sqrt(3) = 6 + sqrt(3) (that's (A.B))

then put that into the projection eq. to obtain ((6+sqrt(3)/(5))V).

And that is my answer, but it just seems a little funky. Could you check to see where I've made an error? If something is confusing, just let me know and I'll clarify, half of this is a description of a diagram.

Last edited: Sep 2, 2008
2. Sep 4, 2008

### Defennder

The question certainly is really confusing. You mentioned it came with a diagram. That would help a lot if you post it.

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