Let it therefore be made to touch a right cone, vertex in any point mm in PM and generatrix m'ß making an angle of degrees with its base gko. The horizontal trace of this plane must touch gko; (Theorem VII.).

And since the plane is to contain PM, the horizontal trace rp of the plane must pass through p, (Theorem VI.).

Through the point mm' draw the horizontal line mn, m'n' parallel to the trace rp, and determine its vertical trace nn'.

Through n' and s draw the vertical trace of the plane.

Observe that the plane rsn by containing PM is perpendicular to plane ; and, by touching the cone described, has the required inclination ; it thus satisfies the conditions of the problem.

Note also that the inclination of PM is 90-0; since if a straight line and an inclined plane be perpendicular to each other their inclinations are complementary.

Cor. To determine three planes each perpendicular to the other two.

For the first and second, determine planes qlm' and rsn' as in the above problem.

Find their intersection mq, m'q' by Problem 12.

The plane which is perpendicular to MQ is perpendicular to the planes which meet in this line (Eucl. xI. 19). Therefore such plane, determined by Problem 13, is the 3rd plane required.

PROLLEM XXVI.

To determine a plane inclined at an angle of 0 degrees and making a given angle a with a given inclined plane.

Let qlm' be the given inclined plane.

To fulfil both conditions, the required plane must touch two right cones having a common vertex in any point vʊ

with circular bases of and b's' in the horizontal and inclined planes respectively*; the generatrix of the former making an

b' is the point of intersection of the lines v'r and Im' and is not lettered in the fig.: also the plan of the circular base b's' is not shown in the fig.

angle

with its base, and that of the latter an angle of a degrees with the inclined plane.

The horizontal trace pt of the common tangent plane to the cones is (Theorem VII.) a tangent to the horizontal traces of the two conic surfaces, i.e. the trace pt touches the circle op and the ellipse rtu. Since four such lines can be drawn, as appears by the figure, it is obvious that in the case. here indicated four planes can be determined which satisfy the given conditions of the problem.

Should the circle op fall within the ellipse, without touching it, the given conditions are impossible.

PROBLEM XXVII.

Given three lines ag, ak, ar meeting in a point a; as the plans of the straight lines AG, AK, AR, each of which is at right angles to the plane of the other two; to determine an elevation of the lines.

Since the plane of lines AK, AR is perpendicular to the line AG (Eucl. XI. 4), the horizontal trace, cd, of the plane of AK, AR is perpendicular to ag, the plan of AG; (Theorem V.).

Therefore draw cd at right angles to ga produced, and meeting ak, ar in points c and d.

ce-drawn through c at right angles to ar, and de drawn through dat right angles to ak, will meet ag in e and be the horizontal traces of the other two planes; (Theorem V.).

Take the projecting plane of the plan of AG for a plane of elevation, and let it revolve about the line ag as an xy.

This plane cuts the plane of AK and AR in AP or a'p, which is therefore perpendicular to AG or a'g in the plane of elevation; (Eucl. xi. def. 3).

d

Fig. 27.

Therefore on ep describe the semicircle cap, through a draw aa' perpendicular to xy, and ea', ap are the required elevations of the three lines; (Theorem III. 2).

PROBLEM XXVIII.

Given two points B and C at unequal distances from a given plane, to determine a point Q in the plane at which the straight lines QB and QC shall make equal angles with the plane.