I am stuck on the following question, which I think is relying on a basic understanding of vectors and geometry applied in a different way.
If $O = (0, 0, 0), A = (1, 0, 0), B = (1, 1, 1)$ and $C = (0, 2, 0)$, find $\underline S$, the vector
surface area of the loop OABCO by:i. drawing the loop projections onto the yz, zx and xy planes and thus finding the
components of s; andii. filling in the loop with two or three plane polygons, finding the vector area of
each, and combining the results.
I am stuck on part i). If I first consider the yz plane, I have three distinct points to work with: $(0,0),(1,1),(2,0)$
I do not see how I can see the porjection of the loop on the plane from these three points, let alone find its area! Is this projection not an ellipse? If so, I do not think three points are sufficient to define the ellipse…
Perhaps I am thinking of this wrong and simply taking the y and z components of each vector does not give the projection of the loop onto the yz plane?
Best Answer
The three points you list $(0,0,0), (0,1,1)$ and $(0,2,0)$ do not form an ellipse. They form a triangle in the $yz$ plane as shown in the figure below, where the B and C in that figure are the projections of the $B$ and $C$ in the question. You can find the scalar area of this triangle and then attach an appropriate unit normal (which will be $\pm\hat{\mathbf{i}}$ ) and this is the $x$ component of $\mathbf{S}$. You can find the $y$ and $z$ components of $\mathbf{S}$ similarly. Finally $\mathbf{S}=({S}_x,{S}_y,{S}_z)$. Note, not all projections will necessarily be triangles, but they will all be simple planar figures whose scalar area can easily be determined.