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parallel to these traces, and the horizontal projection mn, of the intersection will be also parallel to them, and its vertical projection m'n' parallel to the ground line.
3o. Let the two planes meet the ground line at the same point; and let the two planes be apa' and bpb'. The general construction will not apply to this case. must be cut by another plane, and the projections of the two lines of intersection found. The points where these projections meet will be the projections of a point common to the two planes; and as the point p, where these planes meet the ground line, is also common to them, it follows that the projections of their intersection are completely determined.
In general, the third plane of projection is taken perpendicular to the ground line, the traces of the given planes upon it are determined, and thus the construction is referred to the general case. This solution is that given in the figure.
The traces of the third plane are aa, aa', both - to xy: they meet the traces of the plane ap'a in a and a'; and, if we conceive the vertical plane of projection to be
placed, for an instant, in its true position, the line joining a and a' will be the trace of the plane apa' upon the third plane. Now make this third plane turn round aa till it coincide with the horizontal plane; the point a will not be changed, aa' will coincide with xy, and the point a will coincide with a" if aa" =ad', consequently, the line joining a and a' in space, will coincide with aa”.
In the same manner the trace bb” of the other plane upon the third plane may be found; the point d, where it cuts aa", is common to the two planes, and we have merely to determine the projections of this point upon the primitive planes.
Now, if d m be drawn 1 to a a, its horizontal projection will be m: and if d d be drawn I to xy, and we take in aa', am' = dd', its vertical projection will be m'. Join :: pm and pm': these will be the projections of the intersection required.
4°. Let the planes be parallel to the ground line, then their traces and their intersection are also parallel to that line, and the general construction does not apply. We must in this case also use a third plane, as we have done above ; but the mode of proceeding will be precisely the same as that above described, and will therefore be sufficiently intelligible to the learner without further description.
To find the intersection of a line with a plane.
17. Let a plane pass through the given line and cut the given plane: the point where the given line and the line of the planes' intersection meet will be the point required.
1°. Let the plane passing through the given line be vertical, that is 1 to the horizontrl plane.
Let aa, a'a, be the traces of the given plane; bcb'c' the projections of the line. Produce bc to m, draw mm' 1 to
xy, meeting aa' in m'; then bm and mm' will be the hori
zontal and vertical traces of the B
plane through the given line; let 6 m meet a a in d: then d and m' are the intersection of the traces of the two planes ; and if dd be drawn 1 to xy, and m'd
be joined, m'd' will be the ver8
tical projection of the intersec
tion of the two planes. The vertical projection of the required point should be on m'di but it must also be upon the vertical projection B'c' of the given line; it is :. at o', the point of their intersection.
The horizontal projection o may be found by drawing d'o+to ay, and meeting bc in o.
The point o may be found as we have determined ó', by supposing the third plane to be 1 to the vertical plane. Its traces are n'n and n'b', and they cut those of the given plane in n and e; then draw e'e + xy, join en, it must meet bo in the point o.
29. Let the plane passing through the given line have any position whatever.
It is obvious that its traces will pass through those of the given line: first, therefore, determine the traces, n and m' of the given line, bc, b'c'; then through those traces, and any point B in xy, draw the lines Bp and Bp'. The given line is in the plane pBp', since two points of it are in that plane.
Next, to determine the intersection of the two planes. From d and d', where the traces of the planes intersect, draw dd' and e'e 1 to wy, draw the lines de and d'e, these are the projections of the intersection of the two planes.
Now the projections of the required point ought to be in these projections and in those of the given line, bc, b'd': hence the horizontal projection ought to be at o, the vertical projection at o', and .. the line oo' should be I to xy.
Cor. Suppose the given line to be 1 to one of the planes of projection,- for instance, the horizontal plane, -its horizontal projection will be a single point o, and its vertical projection oʻo" is 1 to x
XY. The plane cdď passing through it is vertical, and its trace dd 1 to wy; but its horizontal trace has only one condition, namely, that of passing through the point o.
The other part of the construction is the same as in the general case, the vertical and horizontal projections of the required point being o' and o.
The preceding construction will enable us to solve the following problem :-Given one of the projections of a point situated in a given plane, to find the other projection.
To find the points of intersection of three given planes.
18. The planes, taken two and two, give three lines, which pass through the required point. We therefore by Problem 2 find the projections of these lines; and, when the constructions are correctly made, the three horizontal projections intersect in the same point, which is the horizontal projection of the required point; the three vertical projections also intersect in the same point, which is the
vertical projection of the same point; and, lastly, the line that passes throught hese two projections ought to be 1 to the ground line. In the figure, the horizontal projections of the intersections of the three planes, taken two and two, are ad, bf, ce; the vertical projections are a'd', b'f", d'é; and the required point has for its projection o and o'.
PROBLEM 5. Given the projections of two points, to find the projections of
the line which passes through them, and the distance between them.
19. Let a a' and b b' be the projections of the points a and B. Draw the lines ab a'b'; these, it is evident, will be the projections of the straight line AB.
Next, to find its magnitude. We know that aa' and bb' are + to xy, also that ra' and sb' are equal to the altitudes of A and B above the horizontal plane; if :. we conceive two vertical lines to be drawn, one from a and the other from b, and respectively equal to ra' and sb', their extremities will be the points A and B, and the distance between them is the distance required.