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PROPOSITION III. Through a given point to draw a plane perpendicular to a given line. 1. Let AB be the given line, and P the given point situated in AB.
Through AB draw any two planes AM, AN; and in these PC, PD perpendicular to AB. The plane through CP, PD is that required
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2. Let AB be the given line, and P the given point without it.
Through AB draw two planes AM, AN, the former of which passes through P and the latter any whatever; in the plane AM draw PC perpendicular to AB meeting it in C; and in the plane AN draw CD perpendicular to AB. The plane through PC, CD is that required.
Through any point C in CD draw CK parallel to AB; and from any point E in AB, a line EF perpendicular to the plane KCD; draw FG in this plane parallel to CK meeting CD in G; and in the plane EFG draw GH parallel to EF. This line will meet AB, and be perpendicular to both the lines AB and CD.
For since AB and FG are both parallel to CK, they are parallel to one another (Pls. 1. 6); and hence in the same plane. Wherefore also EF is in that plane; and GH, being parallel to EF, is also in that plane, and consequently meets the line AB.
Again, because EF is perpendicular to the plane KCD, and GH parallel to EF, the line GH is also perpendicular to the plane KCD (Pls. 11. 14); and hence it is at right angles to the line CD, which passes through G in that plane.
Lastly, since EF is perpendicular to the plane KCD, the angle EFG is a right angle ; and since AB is parallel to GF, and HG to
EF, and one of the angles EFG a right angle, the figure HF, is a rectangle. The angle ĞHE is hence a right angle.
The line GH is therefore perpendicular to both the given lines AB, CD as was required.
PROPOSITION V. · Through a given point to draw a plane parallel to two given lines.
Let AB, CD be the given lines, and P the given point.
Through P draw PE, PF respectively parallel to AB, CD: then the plane through these is the one required.
For, since AB is parallel to a line PE in c. the plane EPF, it is parallel to that plane (Pls. 1. 3); and similarly, CD is also parallel to that plane. Or, which is the same thing, the plane EPF is parallel to the given lines AB, CD.
PROPOSITION VI. Through a given point to draw a line parallel to two given planes.
Let A be the given point, and MN, PQ the given planes.
Let PN be the common section of the given planes; through A and PN draw the plane APNB; and in this plane draw AB parallel to PN.
Then AB is the line required.
The truth of this follows from (Pls. 1. 9).
PROPOSITION VII. Through a given line to draw a plane perpendicular to a given plane. Let AB be the given line, meeting
B : the given plane MN in A; to draw a plane through AB perpendicular to MN.
Take any point C in AB, and draw CD perpendicular to MN: then the plane through AB, CD is that required.
Through a given line to draw a plane to make a given angle with a
Let AC be the given line, meeting the given plane in A, and CD a perpendicular to this plane. Join AD, and make the angle CED equal to the profile of the given angle; describe the circle EF with D as centre and DE as radius; and draw the tangent AF to it: then the plane CAF is that required.
For join CF, FD. Then the two triangles CED, CFD are right angled at D, and have the sides about the equal angles common or equal. Whence CFD = CED. But CFD is the profile angle of the planes MN and ACF (Pls. 11, 13, and Def. 6); and is hence equal to the given profile angle.
Through a given point in one of two oblique planes to draw a line which
shall make a given angle with the other plane.
Let A be a given point in one u MN of two oblique planes ; it is required to draw a line from A in MN which shall make a given angle with the other plane PQ.
Draw AB perpendicular to PQ, and BC in any direction in the plane PQ; make the angle ACB equal to the given angle; with B as centre and BC as radius, describe the circle DCD', cutting PN (the intersection of the planes) in D and D'. Then AD, AD' both fulfil the conditions.
The proof is obvious and immediate.
PROPOSITION X. Two straight lines in space being given, and a point in one of them :
it is required to draw through the given point a line to meet the other given line, and make equal angles with them both.
Let AB, CD be the given lines, and P the given point in AB; it is required to draw from P a line to make equal angles with AB and CD.
Find GH perpendicular to AB, CD A (Prop. iv.); make GK and GK' each equal to HP; and join PK, PK'. Each of these lines is equally inclined to AB and CD.
For, since HG, HP are equal to HG, GK, and the included angles GHP, PGK are right angles (Constr.), the base HK is equal to the base PG.
Also, since the two sides HP, HK are equal to the two KG, GP, and the base KP common, the angles KPH, GKP are equal. And these are the angles stated in the enunciation to be equal.
In the same manner it may be shown that K'P makes equal angles with AB and CD,
PROPOSITION XI. To construct a trihedral angle equal to a given one. This problem branches out into several cases according to the essential data by which the pattern trihedral is given. In all cases, however, it is possible, in virtue of the theorems already given, to completely construct the trihedral angle or its supplementary one from adequate data.
Suppose this done, we may copy this trihedral angle by taking the three most convenient of its six parts (three plane angles and three dihedral angles) as one data.
The student should be required to discuss the several cases in order.
CONSTRUCTIONS RELATING TO THE PLANE, LINE, AND POINT.
PROPOSITION I. The projections of a point on two of the coordinate planes are given to
find its projection on the third.
Let A be the given point, and YX, XZ, ZY the three planes of projection; there are given a, a, the
Eid. projections of A on the planes XY, XZ to find ag on YZ.
Now in the eidograph, the three lines Aa, Aag, Adz are perpendicular to the three planes; and hence the planes through them two and two are parallel to the coordinate planes and the parallel edges are equal, that is,
aza, = Ou, = azaz = Aa,
Now if we make the plane ZY revolve about OZ till it coincides with XZ, behind OZ; then the line OY will
11 Orth. Elevation and Profile.
Orche coincide with OX, and agu, with a's the points dag as describing quadrants during the revolution.
Whence from 0 in the orthograph as centre, describe the quadrant aza', and draw a'a', perpendicular to OX, meeting the line qa, in a's: this is the orthograph with respect to the elevation and profile.
If with YZ and XY, we proceed in a similar manner, we shall get the orthograph of the point in respect of the plan and profile.
Orth. Plan and Profile.
PROPOSITION II. The orthograph of a line on two planes is given to construct the orthograph of either of them with the third plane.
It is evident that since a line is given when two points in it are given, that if we take the projections of any two points in it, and employ the preceding proposition, we shall only have to draw the line through the two projections upon the third plane to resolve the problem.