Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?
Hint. Set up the line as the diagonal of a box whose edges run along the three axes.
Set the situation up in coordinates. Take a line segment in space and let its ends differ by a along the length axis, b along the depth axis and c along the height axis. The segment is then the long diagonal of a box with edges a, b and c, and its true length is l = √(a² + b² + c²) (the theorem applied twice).
Now read off the three views. Each view flattens one of the three directions: · Front view (on the vertical plane) keeps length and height, loses depth → length √(a² + c²) · Top view (on the horizontal plane) keeps length and depth, loses height → length √(a² + b²) · Side view (on the side plane) keeps depth and height, loses length → length √(b² + c²)
The relation. Add the squares of the three view-lengths: (a² + c²) + (a² + b²) + (b² + c²) = 2(a² + b² + c²) = 2l²
So for any segment,
(front view)² + (top view)² + (side view)² = 2 × (true length)²
A worked check. Take a = 3, b = 4, c = 12. Then l² = 9 + 16 + 144 = 169, so l = 13. Front = √(9 + 144) = √153, top = √(9 + 16) = 5, side = √(16 + 144) = √160. Squares: 153 + 25 + 160 = 338 = 2 × 169 ✓
Two things this makes obvious. · Every view is at most as long as the segment itself, since each drops one positive term. A projection can shorten a line but never lengthen it. · A view equals the true length exactly when the segment is parallel to that plane — for instance the front view equals l when b = 0, meaning the segment has no depth component at all.
✦ Yes. If the segment differs by a, b, c along the three axes, the three views have lengths √(a²+c²), √(a²+b²) and √(b²+c²), so the sum of their squares is twice the square of the true length. Each view is never longer than the segment, and equals it exactly when the segment is parallel to that plane.
