#What are the steps here? Seems complicated.
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deep narwhal
should post it in #1020426321261756536
deep narwhal
well i think i figured it out
if (u1 , u2 ,u3) is a vector its projection on xy plane is (u1, u2, 0) or u1 i+ u2 j thats quite obvious
divide the shadow in three paprallelograms
area of parallelogram with edge vectors as a,b is a x b
now you know the projections
u1 i +u2 j, v1 i +v2 j, w1 i +w2 j
area oa parallelogram = (u1 i +u2 j) X (v1 i +v2 j)
since iXi=0, jXj=0, iXj=K, jXi=-k
area= (u1v2-u2v1)k
similarly, we have remaining areas as(v1w2-v2w1)k, (u1w2-u2w1)k
if we add them the shadow's area is
((u1v2-u2v1)+(v1w2-v2w1)+(u1w2-u2w1))k.......(1)
now we know u X v =w
(u1 v2 - v1 u2)=w3
we get that from cross-product using det and comparing components
from v=wXu
u1w2-u2w1=v3
and from u=vXw
v1w2-v2w1=u3
so our shadows are is
(u3+v3+w3)k
which is exactly the z axis projection by magnitude
SN=u+v+w
its z coordinate is u3+v3+w3
thats it!
@radiant shoal
neat sandal
quite obvious