We are going to place point B in the dihedral system, given by coordinates 1, minus 3 and 2.

The first corresponds to the x axis, the second to the y axis, and the third to z.

On the positive x axis, we measure a unit.

We observe that the second coordinate corresponds to the y axis but negative, therefore it must be above the earth line.

Perpendicular to this line, a segment of three units is drawn.

When drawing the y coordinate, you must put the name of your projection, which in this case will be B one.

To draw the last coordinate, we measure two units in the direction of the positive z-axis, and write the name of its projection of B two.

We see that the two projections of the point are on the same side, and above the line of land. For this reason the point will be located in the second quadrant.

To verify this, we must untie the horizontal plane of projection, and pass it to the profile or profile plane view.

We verify that point b, is located in the second quadrant.

Now, we are going to place point C in the dihedral system, given by coordinates 2, minus 1 and minus 2.

On the positive x-axis, we measure two units.

We observe that the second coordinate corresponds to the y axis, but negative, therefore, it must be above the earth line.

Perpendicular to this line, we draw a segment of a unit, and in its completion, we write the name of the horizontal projection of the point, which is c one.

the third coordinate corresponds to the z axis, but it is negative, so we draw two units below the earth line, and write the name of its projection that is C two.

To find the quadrant where this point is located, we must make a sketch of the profile view, and on it we place the y, y axes.

Subsequently, we place the coordinates corresponding to these two axes.

They are joined perpendicular to their projection planes, and we draw the point in space. Checking that it is located in the third quadrant.

View of the dejection of the horizontal plane of projection, in perspective; and point C in the third quadrant.

Finally, we will place in the Diédrico system the point D, given by coordinates 2, 2 and minus 1.

on the positive x-axis, we measure two units.

We note that the second coordinate, corresponds to the y-axis positive, therefore, must be below the ground line.

perpendicular to this line, a segment of two units is plotted, and in its termination, we write the name of the horizontal projection of the point, which is D one.

The third coordinate, corresponds to the z axis, but it is negative, so, we draw a unit below the ground line, and we write the name of its projection which is D two.

We make a sketch in the profile view, and in it we can see in which quadrant this point is.

We place the Y-axes, and Z. Then we put the coordinates for these two axes.

They bind perpendicularly to their projection planes, and we draw the point in space. Checking that it's located in the fourth quadrant.

View of the dejection of the horizontal projection plane, in perspective.

Point D in the fourth quadrant.