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Now, the simplest way to obtain this view is to make use of the assumed VP, previously drawn in No. 2, touching the corner 3' of the cube. As No. 5 is a view of the cube directly at right angles to that at No. 2, this assumed VP will be behind it in No. 5, and a plan of it in that position will be a line in the HP parallel to the IL. Draw such a line as x y, No. 6 ; then all the points and lines seen in No. 2 from above can be set off from this line. The back lower corner of the cube is touching it, and we know its position is behind 1', No. 5 ; therefore let fall first a vertical projector from this point to x y, No. 6, in point 3', and similar projectors from points 2' and 4'. Set off from x y on the projector through 3', and similar projectors from points 2' and 4'. Set off from x y on the projector through 3', the distance 3' 3 equal to that of point 3, No. 2, from the assumed VP, and on those let fall from 2' 4', No. 5, in points 2' b, 4' b, the distances that their corresponding points in No. 2 are from the same assumed VP. Join points 2' and 4' with 3' by right lines, and parallel to them draw two others indefinitely through point 3. The plan of the line c'c, No. 5, is next required. This may be found in two ways : Either by setting off from xy on the projector through 3', the distance that c in No. 2 is from the assumed VP, and drawing a line through the MECHANICAL AND ENGINEERING DRAWING 131 points so found parallel to xy to cut those drawn through 3 in c'c, No. 6 ; or by letting fall projectors from the ends c c of the corre- sponding line in No. 5 to cut the same lines from 3 as before. The plans of the two short inclined edges of the section between c'b' and cb in No. 5 are found by joining the corresponding points in No. 6 by right lines. On cross-lining the parts of the cube exposed by the