Geometry in three dimensions
Distance. Area. Perpendicularity.
You will use coordinates to measure distances, review a shade sail and choose the position of a right-angle joint. Each problem asks you to connect a diagram with a calculation.
Work in pairs, then complete the checkpoint and exit task individually. Keep labelled sketches, calculations and explanations on paper. Use your GDC in degree mode. Open each answer only after attempting the task.
A flat triangular sail above level ground.
Read the third coordinate
A point in space has three coordinates. Here x and y locate its position across the ground, and z gives its height above the ground. All coordinates are in metres.
P = (1, 1, 2)
Sketch P and write the coordinates of the point directly below it on the ground, where z = 0. This point is P’s ground projection.
If only z increases, what moves?
What stays fixed?
Make your prediction, then select Raise P. Use Reveal ground projection to check which coordinates stay fixed.
Ground plane: z = 0. Coordinates are in metres.
Same planar distance. Same length?
A straight cable connects P to either Q or R. Q and R are directly above the same ground position, but R is higher. All coordinates are in metres.
P = (0, 0, 1)
Q = (3, 4, 1)
R = (3, 4, 13)
Predict which connection needs more cable. Calculate PQ and PR, showing the right triangles you use.
The planar distance is the distance between the ground projections. Why does equal planar distance not give equal cable lengths?
After calculating, select “Reveal next triangle” twice. Mint shows the x- and y-changes; gold shows the vertical change.
Build the distance rule
The cable calculation used two right triangles. Now replace the numbers with coordinate differences to obtain a rule for any two points.
Let P = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂).
Use Pythagoras twice to write PQ.
Let Δx = x₂ − x₁, Δy = y₂ − y₁ and Δz = z₂ − z₁.
- Use Δx and Δy to find the planar distance.
- Combine that distance with Δz to find PQ.
- Write one formula and explain why it works.
The planar distance becomes a side of the second triangle.
Create another point
The previous cable PR was 13 m long. You will now create a different endpoint that uses exactly the same length of straight cable.
Keep P = (0, 0, 1) fixed.
Choose a new point S so that:
- PS = 13 m;
- its vertical change differs from 12 m;
- S lies above the ground.
Record the coordinates of S and substitute into the distance formula to verify PS = 13. Compare with your partner, then find a second example.
Keep the length. Change the components.
The symbols Δx, Δy and Δz are the changes from P to S, not S’s coordinates. Since P has height 1, the height of S is 1 + Δz. Try exchanging component sizes first.
Where is halfway?
A marker is to be placed halfway along the straight cable from P to R. Its position must be halfway in all three coordinates.
P = (0, 0, 1)
R = (3, 4, 13)
Find the midpoint M of PR.
Calculate M. Then write the ground projections of P and R and find their midpoint M′. Explain how M and M′ are related.
Record both sets of coordinates before selecting Reveal midpoints. Explain whether the cable’s midpoint lies on the ground.
Check independently
R = (−1, 2, 4) S = (5, −1, 6)
Work on your own. Write the three coordinate differences, calculate RS in metres and find the midpoint. Enter your results below, then choose an explanation.
Need a worked example?
Try the checkpoint first. If you need support, open the two-step example, then attempt its follow-up before returning to the checkpoint.
P = (1, 2, 2), Q = (4, 6, 14).
Review the proposed sail
A school is considering a triangular shade sail above a seating area.
You have been asked to calculate the edge lengths, the angle at A and the fabric area. You will then check whether the proposal meets the supplier’s limits.
A = (0, 0, 3)
B = (6, 0, 5)
C = (3, 2, 3)
A, B and C are attachment points at the tops of three vertical posts. Coordinates are in metres and the ground is z = 0. Treat the fabric as a flat triangle with straight edges, ignoring sag and stretching.
Each attachment lies vertically above its post base.
Find the three edges
The three attachment points form the sail’s corners. Each fabric edge is the straight-line distance between a pair of points.
A = (0, 0, 3)
B = (6, 0, 5)
C = (3, 2, 3)
Calculate AB, AC and BC.
For each edge, subtract the coordinates and use the 3D distance formula. Draw triangle ABC separately and label its true lengths. Keep exact values for the angle and area calculations.
Calculate first. Highlight the edges to match them to your sketch, then turn to the face-on view to check your labels.
Find the angle at A
You now know all three side lengths of the sail. Use them to find the interior angle where edges AB and AC meet.
AB = √40, AC = √13, BC = √17 metres.
Find ∠BAC using the cosine rule.
Mark A on your sketch and identify its opposite side. Rearrange the cosine rule, substitute the three lengths, then use inverse cosine in degree mode. Give the angle to 3 significant figures.
Two routes to the same angle
SL · What changes the angle?
Imagine opening the angle at A while keeping its two adjacent sides fixed. Sketch two possible triangles, then reason from the cosine rule.
Keep AB = √40 and AC = √13 fixed.
- As BC increases, what happens to A?
- What length of BC makes A = 90°?
- Explain using the cosine rule.
HL · Use the dot product
v = AC = (3, 2, 0)
Both vectors start at A and follow a sail edge. Find their dot product and magnitudes, then calculate the angle between them. Compare your expression with the cosine-rule calculation.
Why must the two methods agree?
Why the formulas agree
Let u = AB and v = AC be the two vectors starting at A. We will connect the dot product to the cosine rule for this same triangle.
Write the vector from B to C in terms of u and v. Try expanding its squared magnitude before revealing the steps.
BC = v − u
|v − u|² = (v − u) · (v − u)
= |u|² + |v|² − 2u · v
BC² = AB² + AC² − 2AB × AC cos A
u · v = |u||v| cos A
A → B → C and A → C have the same displacement. Read the explanation below as you reveal each step.
Select “Reveal next step” to compare your reasoning. The vectors describe the same three sides used in the cosine rule.
How much fabric?
The school needs the area of the flat triangular sail before ordering fabric. Use the two edges meeting at A and the angle between them.
AB = √40 m AC = √13 m
cos A = 18/√520
Calculate the triangular fabric area.
Write the area formula and substitute the values. Keep the unrounded angle on your GDC and give your area in m². Then reveal the perpendicular height and explain how the formula relates to ½ × base × height.
The height to AB is AC sin A.
Area from a cross product
The cross product provides another way to calculate the same sail area directly from the two edge vectors.
u = AB = (6, 0, 2)
v = AC = (3, 2, 0)
Find u × v, calculate its magnitude and halve the result. Compare with your trigonometric area calculation.
Complete the parallelogram in the diagram. Use its two congruent triangles to explain why the magnitude must be halved.
SL: explain how the perpendicular height produces the formula ½ab sin C.
Does the proposal meet the limits?
BC ≈ 4.12 m Fabric area = 7.00 m²
The supplier has two limits for this order. Reveal them, then decide whether the proposed sail meets both.
Supplier limits
BC must be at most 4 m.
Fabric area must be at most 8 m².
Write a separate comparison for the edge and the area, followed by your overall verdict.
Give a separate conclusion for each limit.
Will the shadow also be 7 m²?
The school wants shade on the ground. The fabric area alone may not tell us how much ground is shaded.
The fabric area is 7 m².
Assume sunlight travels vertically downwards onto level ground. Predict where each corner’s shadow falls.
Use A = (0, 0, 3), B = (6, 0, 5), C = (3, 2, 3).
Write the three shadow coordinates, then sketch the ground triangle and calculate its area. Reveal the rays to check your projection. Explain why the shadow area differs from 7 m².
Parallel rays · level ground · vertical sunlight.
Move the sun
Will the seating area stay shaded?
The outlined rectangle is the seating area. The sail stays fixed while the sunlight direction changes. First predict the shadow’s movement; then select Oblique sun or move the slider.
- Compare the shadow’s position and shape with the vertical-sun case. Which parts of the seating area are shaded?
- Record one observation and one piece of information needed to judge shade at the times people use the seats.
Outlined rectangle: seating area. Rays remain parallel.
Which joint makes a right angle?
A frame has two straight supports meeting at a joint R. Their other ends, P and Q, are fixed. The joint can move vertically while its x- and y-coordinates stay fixed. Coordinates are in metres.
P = (0, 0, 1)
Q = (4, 0, 3)
R = (2, 1, h)
The joint can be at h = 2 m or h = 4 m.
Which gives ∠PRQ = 90°?
Use the buttons to compare the two positions. Choose a method and calculate to justify your decision; a right angle may not look square in this view.
The angle to test is at the movable joint R.
Establish perpendicularity
Everyone · Compare squared lengths
P = (0, 0, 1)
Q = (4, 0, 3)
R = (2, 1, h)
Calculate the three squared lengths for h = 2 and h = 4. A right angle at R makes PQ the hypotenuse, so test whether PR² + QR² = PQ². State which position works and why.
HL · Use the dot product
Form vectors RP and RQ, both pointing away from the joint R. Set their dot product equal to zero to establish perpendicularity.
Leave h as a variable and solve your equation. Can any other height work if 1 ≤ h ≤ 5? Explain which solutions are allowed.
Choose a method—and explain why
For each situation below, name a suitable method and explain which given quantities it uses. Work with your partner: one explains, and the other checks the choice.
You know three sides
Find an interior angle.
You know two vectors
Test whether they are perpendicular.
You need a triangle’s area
Choose the quantities you need.
Write a formula for each choice. Explain how the information you have determines the method; then compare your answers with the reveal.
A fresh triangle
Work independently on this new triangle. Use what you have learned to choose a method without opening the answer. Coordinates are in metres.
D = (1, 1, 2)
E = (5, 1, 4)
F = (2, 4, 3)
- Find DE and the midpoint of DE.
- Find the side lengths or vectors you need for ∠EDF. State your method and substitute the values into its formula.
- Calculate the angle if time permits.
HL: explain the effect of reversing just one of the two vectors used for the angle.
Keep your individual evidence
Keep a labelled sketch and full working on paper. Use this box to record your results and explain your method.
Keep your work
Save your checkpoint and written responses.
The download includes your checkpoint attempts and anything typed into the response boxes. It does not include work written on paper, so keep both together.
Entries are saved in this browser when available. Nothing is sent anywhere.
Choose one connection on the right. Explain it to your partner without looking at a formula, then identify anything you still need to practise.
Three connections
Coordinate differences → straight-line distance
Cosine rule ↔ dot product
Triangle area ↔ cross-product magnitude HL
Find every allowable height
The original sail failed the edge limit. Explore whether changing only the height of C can make the design meet both supplier limits.
Return to A = (0, 0, 3), B = (6, 0, 5), C = (3, 2, h), with 2 ≤ h ≤ 8.
Find every height satisfying both limits.
On your GDC, graph each quantity against its own limit. Find the intersections, test which side of each boundary is allowed, then take the overlap. Verify the supplied area model at two heights using triangle methods.