" Were it necessary to drop a perpendicular from the given point upon the right line. we should construct, as has just been described, the coincidence of the right line with the plane drawn by the given point. and which would be perpendicular to it ; and we should obtain. from each of the two projections of the required perpendicular. two points through which it must pass.
" Fifth. S. Two planes being given in position, by means of their traces A B and A L for one, and c D and c d for the other : to construct the projections of the right line upon which they intersect each other.
" Solution. All the points of the trace A B being found in the first of the two given planes, and all those of the trace c D being found in the second, the point E, of intersection of these two traces is evidently in the two planes. and is consequently one of the points of the required right line. In like manner, the point, F, of intersection of the two traces upon the vertical plane, is also another point of this right line. The intersection of the two planes is therefore so placed as to meet the horizontal plane in F. and the vertical plane in F.
It; therefore, the point F be projected on the horizontal plane, which may he performed by dropping the perpen dicular Ff on L at. and if the line,/'E be drawn, it will be the horizontal projection of the intersection of the two planes. So, if the point E be projected on the vertical plane, by dropping the perpendicular E e on Lai, and if the right line e F be di-awn, it will be the vertical projection of the same intersection.
" Sixth 9. Two planes being given, by means of the traces A B and A L for the first, and c n and c d for the second ; to construct the angles formed by them.
"Solution. Having constructed, as in the preceding ques tion, the horizontal projection, E 4 of the intersection of the two planes ; by conceiving a third plane perpendicular to them, and consequently perpendicular to their common inter section ; this third plane will cut the two given planes in two right lines, containing between them an angle equal to the one required.
" The horizontal trace of this third plane will he perpen dicular to the projection. a of the intersection of the two given planes, forming with the other right lines a triangle, of which the angle opposed to the horizontal side will be the one required. It remains, therefore. only to construct
this triangle.
" It is quite indifferent through what point of the inter section of the two first planes the third passes; and we are at liberty to mark its trace upon the horizontal plane, at pleasure; provided that it be perpendicular to Ef. Suppose then, the line c 11 be drawn perpendicular to Ef, terminating, 1 in o and n, at the traces of the two given planes, and meet- ' ing Ef, in the point 1; this line wilt. be the base of the triangle intended to be constructed. In fact, let us suppose that the plane of this triangle turns on its base, o it, as on a hinge, to adapt itself to the horizontal plane ; in this motion, its apex, which was in the first instance placed on the intersection of the two planes, continues in the vertical plane drawn through such intersection, because the vertical plane is perpendicular to G II j and N% hen the plane of the triangle is laid down, this apex will be found on one of the points of the line II! It therefore remains only to discover the heights of the triangle, or the extent of the perpendicular dropped from the point t, on the intersection of the two planes.
" But this perpendicular is comprised in the vertical plane drawn from E to if, therefore, we conceive this plane to revolve about the vertical line f in order to apply itself to the vertical plane of projection ; and if we carry fE from/ to from f to i, the line c F will be the extent of the portion of the intersection comprised between the two planes of projection ; and if from the point i, the perpendicular i k be dropped upon this line, it will give the height of the required triangle.
" Hence, by carrying i h from r to K, and completing the triangle G K n, the angle in K will be equal to the angle formed by the two planes.
" Seventh Question.—Figure 10. Two right lines inter secting each other in space, being given by their horizontal projections A D, A C, and by their vertical projections a b, a c; to construct the angle formed between them.