Mathematics notes, chapter 7
Plans and Elevations
Textbook Mathematics Form 3 (DLP), pages 168–197
What this chapter covers
- Identify horizontal, vertical and inclined planes, and name a normal to a plane (7.1)
- Draw the orthogonal projection of a line, a plane shape and a solid on a horizontal or a vertical plane (7.1)
- Compare an object with its orthogonal projections in terms of the length of sides and the size of angles (7.1)
- Draw the plan and the front and side elevations of an object to scale, using solid, dashed and construction lines, in the four-quadrant layout (7.2)
- Synthesise a plan and elevations to sketch the three-dimensional object (7.2)
- Solve problems involving plans, elevations, lengths, volumes and ratios (7.2)
An architect cannot show a whole building in one flat picture, so the building is drawn from the top and from the sides. The Diamond Building (the Energy Commission headquarters) in Putrajaya looks different when it is viewed from different directions, so its design is described with flat drawings made from the top and from the sides. In this chapter you will learn how a three-dimensional object is projected onto a flat plane, how to draw its plan and elevations to scale, and how to build the object again in your mind from those drawings.
| English | Bahasa Melayu |
|---|---|
| orthogonal projection | unjuran ortogon |
| plan | pelan |
| elevation | dongakan |
| normal to a plane | normal kepada satah |
| horizontal plane, vertical plane, inclined plane | satah mengufuk, satah mencancang, satah condong |
| solid line, dashed line | garis padu, garis sempang |
| quadrant | sukuan |
| scale | skala |
| full scale | skala penuh |
7.1Orthogonal projections
Planes and normals
A plane is a flat surface of an object. A plane can be horizontal (level, like a floor), vertical (upright, like a wall) or inclined (slanted, like a roof).
A normal to a plane is a straight line that is perpendicular to the plane. It makes a right angle with every line that lies on the plane.
Example (textbook Example 1, p. 170): naming normals
In a cuboid PQRSTUVW with base PQRS, the line TS rises from S and is perpendicular to the base. So TS is a normal to the plane PQRS at point S. The same line is also a normal to the plane TUVW, but there its foot is T, so it is named ST.The order of the letters matters: the foot of the normal is written last, as in TS, because it meets the plane PQRS at S.
To find a normal to a plane, look for an edge of the solid that meets the plane at a right angle. In a cuboid, the four upright edges are normals to the top and bottom faces.
What an orthogonal projection is
Orthogonal projections are the images formed on a plane when the projecting lines from the object are perpendicular (normals) to the plane.
For a line, drop a normal from each end to the plane and join the two feet. For a solid, do this for every vertex, then join the feet in the same order.
A projection is orthogonal only when the projecting lines are normal to the plane. Slanted lines still make a shadow, but it is not an orthogonal projection (textbook Example 3, p. 172).
Drawing an orthogonal projection
- Decide the type of plane (horizontal or vertical) and the viewing direction.
- Draw normal lines from every vertex to the plane.
- Join the points where the normals meet the plane, in order.
- Redraw the shape with the actual measurements and label the vertices (a vertex hidden behind another is written after a slash, such as E/D).
Example (based on textbook Example 5, p. 176): a cylinder
A cylinder of diameter 4 cm and height 3 cm stands on a horizontal plane.(a) Viewed from above (on the horizontal plane): a circle of diameter 4 cm.
(b) Viewed from the side (on a vertical plane): a rectangle 4 cm wide and 3 cm high.
A cone of base diameter 6 cm and height 4 cm stands on its base on a horizontal plane. What are its orthogonal projections on (a) the horizontal plane viewed from above and (b) a vertical plane viewed from the side?
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(a) A circle of diameter 6 cm, with a dot at its centre (the apex is directly above the centre).
(b) An isosceles triangle with base 6 cm and height 4 cm.
(b) An isosceles triangle with base 6 cm and height 4 cm.
Comparing an object with its projection
The length of a side and the size of an angle in the projection may stay the same or may change, depending on the viewing direction.
- A side that is parallel to the projection plane keeps its actual length.
- A side that is inclined to the plane becomes shorter.
- A side that is perpendicular to the plane becomes a point.
- An angle on a face that is parallel to the plane keeps its size. An angle on an inclined face may become larger or smaller.
Example (based on textbook Example 6, p. 181): a ramp
A right prism has a rectangular base PQRS lying on a horizontal plane, 8 cm long and 2 cm wide, and a vertical rectangular end PSUT 8 cm long and 3 cm high. The sloping face is TUQR.Viewed from above, the plan is a rectangle 8 cm by 2 cm. The lengths PQ and QR are the actual lengths and the right angles stay right angles. The sloping edge TQ is inclined, so its projection (PQ = 2 cm) is shorter than TQ itself, which is √(2² + 3²) = √13 ≈ 3.6 cm.
Do not assume that every length in a projection is the actual length. Only sides parallel to the plane are drawn at their true size.
A 10 cm rod leans so that its top is 8 cm above the ground and its lower end touches the ground. How long is its orthogonal projection on the ground?
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√(10² − 8²) = √36 = 6 cm. The projection is shorter than the rod because the rod is inclined.
7.2Plans and elevations
The orthogonal projection on a horizontal plane, seen from the top, is called a plan. The orthogonal projections on vertical planes, seen from the front or the side, are called elevations (front elevation and side elevation).
Plans and elevations are drawn to full scale (actual size) or to a scale given in the question. In the textbook, the object stands on a horizontal plane, the plan is seen from Z (above), the front elevation from X (in front) and the side elevation from Y (from the right-hand side in method 1, from the left-hand side in method 2).
Line types
If you could see an edge from the viewing direction, draw it solid. If it is hidden behind the object, draw it dashed. Use thin lines only as guides.
Worked example: a house-shaped prism
The prism below has a rectangular base 6 cm by 4 cm, upright walls 3 cm high and a ridge 5 cm above the floor in the middle.
- Plan from Z: a rectangle 6 cm by 4 cm. The ridge is a solid line down the middle, at 3 cm from each side.
- Front elevation from X: a pentagon (a 6 cm by 3 cm rectangle with a triangle of height 2 cm on top).
- Side elevation from Y: a rectangle 4 cm by 5 cm. The line at 3 cm is where the wall meets the roof; it is visible, so it is solid.
The four-quadrant layout
- (Method 1) Draw the plan in the fourth quadrant (bottom right).
- Draw the front elevation directly above the plan (first quadrant), using thin vertical lines from the plan so the widths match.
- Draw a 45° line from the centre point into the third quadrant. Project the plan across to the 45° line, then straight up, and use thin lines from the front elevation across for the heights. This gives the side elevation in the second quadrant (method 1, viewed from the right).
In method 1 the side elevation is seen from the right and sits to the left of the front elevation. In method 2 it is seen from the left and sits to the right; the plan is then drawn in the third quadrant, under the front elevation, and the 45° line is in the fourth quadrant.
Textbook figure: Mathematics Form 3 (DLP) textbook, p. 187 (Example 9).
Hidden edges: a block with a channel
- Plan: both edges of the channel are visible from above, so two solid lines.
- Front elevation: the channel shows as a notch, 2 cm wide and 1 cm deep.
- Side elevation: the channel's floor is hidden inside the block, so it is drawn as a dashed line at 2 cm from the bottom.
Textbook figure: Mathematics Form 3 (DLP) textbook, p. 182.
The block above is viewed from the left (the opposite side to Y). Is the line at 2 cm solid or dashed? What is the volume of the block?
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Still dashed: the channel runs the whole depth of the block and is hidden from either side.
Volume = 6 × 4 × 3 − 2 × 4 × 1 = 72 − 8 = 64 cm³.
Volume = 6 × 4 × 3 − 2 × 4 × 1 = 72 − 8 = 64 cm³.
Forgetting a hidden edge, or drawing it solid. A line that cannot be seen from the viewing direction must be dashed.
Sketching the object from its plan and elevations
Plan and elevations share measurements, so together they fix the shape. To rebuild the object:
- Sketch the three views on their planes using the given measurements.
- Project each surface out from its view so that the matching surfaces meet.
- Join the vertices and label them with the letters of the views.
- Write the lengths on the sketch.
Example (based on textbook Example 9, p. 187): a three-step block
Use the front elevation of the block above to calculate its volume.
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Area of the front elevation = 2 × 3 + 2 × 2 + 2 × 1 = 12 cm². Volume = 12 × 4 = 48 cm³.
Solving problems
Combine the measurements from the views with area and volume formulae. Volume of a prism = area of uniform cross section × length. A triangular piece removed from a block has volume ½ × base × height × length.
Example (based on textbook Example 11, p. 190)
A block was a cuboid 9 cm long, 4 cm wide and 6 cm high. A right prism with a triangular cross section (base 4 cm, height 3 cm) and length 4 cm was cut off one top edge. Find (a) the volume removed, (b) the ratio of the volume removed to the volume left.(a) ½ × 4 × 3 × 4 = 24 cm³
(b) Cuboid = 9 × 4 × 6 = 216 cm³. Left = 216 − 24 = 192 cm³.
Ratio = 24 : 192 = 1 : 8
Check the slanting edge in the front elevation with Pythagoras: 4² + 3² = 5², so the slope is 5 cm. The slope appears with its actual length only in the view where it is parallel to the plane.
The plan of a solid is a circle of diameter 6 cm with a dot in its centre. The front elevation is a rectangle 6 cm wide and 5 cm high with an isosceles triangle of height 4 cm on top. (a) Describe the solid. (b) How long is each slanting side of the triangle?
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(a) A cone of height 4 cm standing on a cylinder of height 5 cm; both have diameter 6 cm.
(b) √(3² + 4²) = 5 cm.
(b) √(3² + 4²) = 5 cm.