PROBLEM XIV. Figs. 39 and 40.

To find the point of intersection of a line and a plane. Let ab, a B be the projections of the line and LMN the traces of the plane.

Construction. Take the vertical plane AaB containing the line AB, and find the vertical projection Cd' of the common section of AaB and LMN. The point o' where a'B and Cd intersect will be the vertical projection of the point required; the other projection will be o on ab.

Proof. Since AB and CD are in the same plane AaB they intersect in the point 0. But CD is also in the plane LMN, and therefore O is in that plane.

Hence AB meets LMN in 0.

Remark. Any plane whatever might be taken containing AB; but when the plane is taken perpendicular to one of the planes of projection, as above, the solution is shortest.

PROBLEM XV. Figs. 41 and 42.

To determine the projections of a line which shall contain a given point and be perpendicular to a given plane.

Let pp' be the projections of the point and LMN the traces of the plane.

Construction. Draw pq perpendicular to MN and p'q' perpendicular to LM; pq, p'q will be the projections of the line containing P and perpendicular to LMN.

Proof. Because MN is horizontal it is perpendicular to Pp, and since it is also perpendicular to pS it is perpendicular to the plane PQS (Theor. III. Cor. 1), and therefore the plane LMN is perpendicular to PQS (Theor. IV.).

It may be proved in a similar manner that LMN is perpendicular to the projecting plane p'PQ.

Therefore, since PQS and p'PQ are perpendicular to LMN, their common section, PQ, is perpendicular to it (Theor. VI.).

Cor. 1, To determine the distance from a point to a plane.

Draw the projections of the perpendicular and find the point of intersection Q by Problem XIV.; then determine the length of the line PQ (Prob. IV.).

Cor. 2. To determine a plane parallel to a given plane and at a given distance from it.

From a point in the given plane draw a line at right angles to it and set off the given distance from that point along the line (Prob. V.); then through the point so found draw a plane parallel to the given plane (Prob. XII.).

Cor. 3. To determine the traces of a plane which shall contain a given line and be perpendicular to a given plane.

From any point of the given line draw a perpendicular to the given plane, and find the traces of the plane containing these two intersecting lines (Prob. x.). This will be the plane required (Theor. IV.).

Cor. 4. To determine the traces of a plane which shall contain a given point and be perpendicular to two given planes.

From the given point draw a perpendicular to each of the given planes, and find the traces of the plane containing these two perpendiculars (Prob. x.).

Note. It has been proved in this problem that when the projections of a line are at right angles to the traces of a plane the line is perpendicular to the plane. The converse is also true; i.e. if a line be perpendicular to a plane the projections of the line are at right angles to the traces of the plane. For, otherwise two perpendiculars might be drawn to a plane from the same point, which is impossible (Theor. III. Cor. 3).

PROBLEM XVI. Figs. 43 and 44.

To determine the traces of a plane which shall contain a given point, and be perpendicular to a given line.

Let pp' be the projections of a point P, and ab, a'b' the projections of a line AB; it is required to find the traces of a plane containing P and perpendicular to AB.

It follows from Problem xv., that the traces of the required plane must be at right angles to ab, a'b'; so that it is only necessary to determine a point in one of the traces: hence the following:

and

Construction. Through P draw a line parallel to the horizontal trace of the required plane, that is a line PQ having its horizontal projection pq at right angles to ab, its vertical projection parallel to the ground line, and find the vertical trace Q of this line. Through Q draw QM perpendicular to a'b', and through M draw MN perpendicular to ab. QMN will be the plane required.

Proof. Because the lines PQ and MN are parallel they are in the same plane; but the point is in the plane QMN, therefore the whole line PQ is in that plane.

That is the plane QMN contains P, and it is perpen

dicular to AB (Prob. xv.).