PROBLEM V. Fig. 83.

To find the traces of a plane which shall contain a given line and have a given inclination.

Let AB be the given line and the given angle; it is required to draw the traces of a plane containing AB and having an inclination to the horizontal plane.

Construction. Take v' the projections of any point on AB. Draw v'c' making the angle Ø with wy. Now suppose VC to revolve about Vv, generating a right circular cone the trace of which is the circle CD. Draw that circle, and from B, the horizontal trace of AB, draw BE a tangent to it at the point D. BEA is the plane required.

Proof. The plane BEA contains the line AB, since it contains the traces A and B. As it contains the vertex V of the cone, its inclination is the angle between VD and vD. (Prob. XIX. Ch. II.)

But the angle VDv=0 by construction.

Therefore BEA is the plane required.

As two tangents to the circle may be drawn from B there are two solutions to this problem.

Corollary. The angle between a tangent plane to a cone of revolution and a plane at right angles to its axis is equal to the angle contained by the hypotenuse and base of the right-angled triangle which generates the cone.

E

Fig. 83.

B

PROBLEM VI. Fig. 84.

To find the traces of a plane which shall contain a given line and make a given angle with a given plane.

Let LMN be the given plane and AB the given line; it is required to find the traces of a plane containing AB and making an angle ◊ with LMN.

The method of solving this problem is suggested by the corollary to the preceding one. For if a cone of revolution be conceived, having its vertex in the given line, its axis perpendicular to the given plane, and its generators making an angle with that plane, then the tangent plane to the cone which should also contain the given line would be the plane required. Hence the following:

Construction. Through any point A of the given line draw a vertical plane DEL at right angles to the given plane. Let DEL be turned into the vertical plane of projection, bringing with it the point A and the line of intersection DL, these are now A, and DL. From A, draw 4,G, at right angles to DL, and AH, making an angle with DL. GH, is the radius of the circular trace of the cone with LMN. Now the line LD, being returned to its original position, and the plane LMN rabatted on the horizontal, the point G comes to G2, and H to H2 This rabatment is most easily effected by producing ED and making DG, equal to D,G,, and DL, equal to DL. With centre G, and radius GH describe the circle HKO, which is the rabatment of the trace of the cone with LMN. Next find the point of intersection of AB with LMN and its rabatment P1, and draw P,Q a tangent to the circle. NQ is the rabatment of the line of intersection of the required plane with LMN.

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NRQ, which contains the two intersecting lines AB and NQ,, is the plane required. The second tangent to the circle gives a second plane whose traces are STU.

If the given line and plane were parallel to one another, the line of intersection of the tangent plane to the cone with LMN, that is the line NQ, in the figure, would be parallel to the given line. (Theor. xi. Ch. 1.)

PROBLEM VII. Figs. 85, 86 and 87.

To find the section of a cone of revolution by a plane.

First case, Fig. 85. Let ABC, a'b'v' be the projections of a cone of revolution, and LMN the traces of a plane which cuts all the generators, but not at right angles to the axis; it is required to find (1) the projections of the curve of intersection of the plane with the cone, and (2) the true form of that section.

Construction. The horizontal plane is taken at right angles to the axis of the cone, and the vertical plane at right angles to LMN. The vertical projection of the required curve is therefore the straight line o'N.

To find the horizontal projection of any point P of the curve, when the vertical projection p' is known; draw, as in Prob. I., the vertical projection v'p' of the generator through P, and determine the horizontal projection Dv of that generator; the point p on Dv is the horizontal projection required.

In the same way any number of points on the horizontal projection of the curve may be found and the curve traced through them.

This method fails for the point q, which is best determined by drawing the horizontal projection of the circle described by the point Q as it revolves about the axis; q'r' is the radius of that circle; therefore the point q is on the circle described about v with the radius vr.

Next, to find the true form of the section, the plane LMN is rabatted on the horizontal plane of projection, which determines the curve N,Q,P,O,S,. That curve is an ellipse.

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The curve opn is also in general an ellipse, though it may be a circle.