If a line is perpendicular to a plane perpendicular to their projections are traces of the plane. The converse is not true, that the projections are perpendicular not mean that the line is perpendicular to the plane.

The converse is always true if satisfied that the projections of the line are perpendicular to the trace of the plane on the three planes of projection, plan, elevation and profile.

All planes through that line is perpendicular to the plane, so that to a plane perpendicular to another just to make a line perpendicular to the plane and then pass a plane through that line, it is determined that the 2 planes are perpendicular. If a line is perpendicular to a plane it is all straight from the plane, so that given a line perpendicular to a line, make a plane perpendicular to it (one whose traces are perpendicular to their projections) and emphasize any line on the plane. The line in the plane is always perpendicular to the given line.

In the left figure, a straight line to a plane perpendicular to both the alpha is all straight from the plane: b, c, d.

In dihedral system, a line is perpendicular to a plane so that their projections r1 r2 are perpendicular to the trace of the plane. In the drawing the line perpendicular incident at the same time by a point A. If an alpha plane is perpendicular to a straight, all lines of that plane (as b) are perpendicular to the line a. So to make a line perpendicular to the straight, just make a plane perpendicular to it and any straight line of that plane.

In the left figure, a straight line to a plane perpendicular to both the alpha is all straight from the plane: b, c, d.

The necessary and sufficient condition for a line is perpendicular to the plane that is perpendicular to 2 nonparallel lines of the plane (as it may be perpendicular to two parallel plane to the plane without being one).

In the middle figure, a straight is perpendicular to a plane alpha and therefore all lines of the plane: b, c, d.

As the line is perpendicular to the plane of its projections a1 a2 traces are perpendicular to the plane.

Any plane passing through a, beta for example, is perpendicular to alpha.

In the right figure any plane passed through the straight beta perpendicular to the plane alpha, beta this plane is perpendicular to affect the alpha line a.

In the middle figure, a straight is perpendicular to a plane alpha and therefore all lines of the plane: b, c, d.

As the line is perpendicular to the plane of its projections a1 a2 traces are perpendicular to the plane.

Any plane passing through a, beta for example, is perpendicular to alpha.

In the right figure any plane passed through the straight beta perpendicular to the plane alpha, beta this plane is perpendicular to affect the alpha line a.

In dihedral system, a line is perpendicular to a plane so that their projections r1 r2 are perpendicular to the trace of the plane. In the drawing the line perpendicular incident at the same time by a point A. If an alpha plane is perpendicular to a straight, all lines of that plane (as b) are perpendicular to the line a. So to make a line perpendicular to the straight, just make a plane perpendicular to it and any straight line of that plane.

To make a plane perpendicular to an oblique line through a point P, is any straight line (in blue) whose projection on the ground passing through P1 and perpendicular to r1, which intersects the ground line is a vertical until it intersects to the horizontal projection of the line by P2. At that point of intersection is the trace of the plane perpendicular to beta 2 where r2 and intersects the ground line is a perpendicular beta1 to R1. Beta plane is perpendicular to the line r and passing through Q.

If a line is perpendicular to a plane it is all the straight plane, the converse is not always true. The necessary and sufficient condition for a line perpendicular to a plane is that it is not parallel to two lines of the plane.

In the drawing can verify that this is true, we have a straight yellow b that is perpendicular to the plane and therefore to all the straight plane, between which are two parallel lines m n. Also observed a red line two parallel lines perpendicular to the plane which however is not perpendicular to the plane. This could be arranged by having these two lines were not parallel with what we should all line perpendicular to two parallel lines do not be perpendicular to the plane containing them. Thus the red line was never confirmed that would not meet the condition of being perpendicular to two nonparallel lines plane and therefore is not perpendicular to the plane.

The projections in plan and elevation of the red line we can verify that for being a straight profile and despite compliance with the condition of perpendicular (that is that the projections of the line perpendicular to the trace of the plane), we note that line is not perpendicular to the plane, so to ensure compliance that the line is perpendicular to the plane in the case of the line profile we will also require the condition of perpendicularity in profile plane, so the three projections of the line would have to be perpendicular to the plane of the three tracks. Thus we can conclude that limitation, if that situation occurs in the three planes of projection, that is, that the three projections of the line perpendicular to the traces of the plane, we have that the line is always perpendicular to the plane, no exceptions .

To construct an alpha plane perpendicular to an oblique plane beta, is a straight perpendicular thereto having their projections therefore a1 a2 perpendicular to their traces. The following is any alpha plane which affects this straight line (in the drawing a vertical plane) and have it is perpendicular to the given plane, and that if we take a line perpendicular to a plane all planes passing through that line are perpendicular plane.

Alpha in a frontal plane (red), ie a plane parallel to the vertical plane, it comes to a straight line to and one beta plane perpendicular thereto.Then draw a line perpendicular to the line b above.

To construct a line perpendicular to a plane in front, must be satisfied that their projections are perpendicular to their traces, so the plan projection is perpendicular to the horizontal trace of the plane and its projection in the profile is at your trace in the profile plane. All lines perpendicular to a plane are straight front edge and that are also perpendicular to the vertical plane of projection. All plane through the line is perpendicular to the given plane, eg the plane that appears in green called flat projecting edge or vertical.Finally, to construct a line perpendicular to the line b above we can draw a vertical line such that impinges on the frontal plane, a plane perpendicular to the line above.

On the right is the exercise solved the dihedral system, the frontal plane represented by its horizontal trace the straight-edge and perpendicular to it. A flat edge that falls in this line and thus is perpendicular to the plane above and a line perpendicular to this straight, vertical line which is incident on the red background.

The figure shows a red background alpha, parallel to the ground, trying to do a line to and beta plane perpendicular to it. It is further proposed to make a line perpendicular to the line b above.

As the plane is parallel to the ground need to build the three projections of the line, because even being projected in plan and elevation of the line perpendicular to the trace of the plane, is not reason enough to confirm that indeed the line is perpendicular to the plane.

The line perpendicular to the plane has its projection (a1 a2 a3) perpendicular to the trace (alpha 1 2 and 3) and a beta plane perpendicular to the die passes through any line perpendicular to it, straight as before. Since we have that if a challenge is perpendicular to the plane it is all straight from the plane, to a line perpendicular to the given line do a line any given plane b.

On the right is at a lower level of exercise solved the dihedral system: the three traces of alpha plane orthogonal projections of the line to indicate that it is perpendicular to the plane, also a red line belonging to the plane indicates that it is perpendicular to the line a.

The figure shows a map of alpha profile in which we will calculate a straight line to and beta plane perpendicular to it.In addition we calculate a line perpendicular to the line b above.

The profile plane is perpendicular to the ground line so that a line perpendicular thereto is parallel to the same line, such as straight. Any plane through the line is perpendicular to the former, such as beta plane parallel to the ground line. If we make a line perpendicular to the previous one draw any plane perpendicular thereto, for example line b.

To resolve the current financial dihedral system, we represent the profile plane (in this case we did match the profile plane that represents the trace of the plane perpendicular to the given plane) and perpendicular to it straight and a beta plane impinges on this line (blue plane defined by the three squares).

To make the line perpendicular to the plane to draw its projections a1 a2 traces perpendicular to the alpha 1 and alpha 2 to a plane perpendicular to the plane had previous beta this line, to thereby represent the profile trace by passing its third beta 1 by the third projection of the straight a3. Finally a line perpendicular to the given b happens for example by the alpha plane perpendicular thereto, thereby the projections of the straight line b1 b2 traces coincide with the plane perpendicular to the line alpha 1 and 2.

Have a flat edge alpha and we construct a line perpendicular to the same, a beta plane perpendicular thereto and a line perpendicular to the first b.

If the line is perpendicular to its projections are perpendicular to the trace, we must therefore a1 is perpendicular to alpha one to alpha 2 a2, a3 perpendicular to alpha 3. All beta plane passing through this line is perpendicular to the plane alpha. We have a frontal plane that has its horizontal trace parallel to the ground (in the drawing in blue). Finally any line b alpha plane is perpendicular to the straight and that if a line to a plane perpendicular to it is at all lines of the plane so simply take any straight line b of this plane to meet such condition.

In the small picture to the right we see the exercise resolved dihedral system, the alpha level of singing black color and the line perpendicular to the same incident in it a plane perpendicular to the given beta. Also observed a straight plane b of red and therefore perpendicular to the straight, perpendicular to the plane alpha.

## No hay comentarios:

## Publicar un comentario