Vectors: lines and perpendicular feet
By the end, you can
- Write a line using a point and direction
- Find intersections and recognise skew lines
- Use orthogonality to find a perpendicular foot
about 30 minutes for the lesson · practise at your own pace.
The central idea
A line has equation , with . A point is on the line only if a single parameter satisfies every component. For the nearest point to P, require . Lines in space can be parallel, intersecting or skew.
Worked example
For and , write . Requiring gives , so . The distance is . Reflection in the line gives . Substitution into the line and the zero dot product verify the foot; the midpoint of P and Q verifies the reflection.
A parameter labels points unless the model defines time. For a foot, check both line membership and perpendicularity; for a reflection, check the midpoint too.
Check your understanding
A light ray starts at and travels in direction towards the plane . Find its landing point and the distance travelled. Explain how you check that the supporting-line intersection belongs to the forward ray. If the direction is reversed, can that forward ray reach the plane?
The ray is with . At , , so and the landing point is . The distance travelled is , not 3: this direction is not a unit vector.
For the reversed direction , the plane condition is , so . This point belongs to the supporting line but violates ; the reversed forward ray does not reach the plane.
Mastery learning objectives
At the end of this chapter, I should be able to:
Choose a practice question
Use a lecture example to review a method, then attempt its linked Discussion and Review questions before opening the working. Part references below follow the printed skill table.
| Skill | Lecture examples | Discussion | Review follow-up |
|---|---|---|---|
| Line equations: vector, Cartesian and parametric | Example 20, Example 21, 27(iii) | 1, 2, 7(i)–(iii), 8(a) | 3(i), 4(i), 7(i), (iv) |
| Whether a point lies on a line | Example 22, 27(iv) | 3(i) | 5(i), 9(i) |
| Foot of a perpendicular | 23(i), 27(v) | 4, 7(ii), first part, 10(iii) | 4(iii), 5(iii), 7(iii) |
| Perpendicular distance from a point to a line | 23(ii), Example 24 | 3(ii), 4 hence, first part | — |
| Relationship between two lines | Example 25 | 5, 6 explain, 9(i), 10(i) | 1 first part, 2(i), 10 second part |
| Angle between two lines | Example 26, 27(ii), internal angle | 5 | 1 second part, 7(ii), 9(iv) |
| Basic vectors: magnitude | — | — | 3(iii) |
| Basic vectors: equal vectors | 27(i) | — | — |
| Basic vectors: ratio theorem | 23(iii) | 4 hence, second part, 7(ii), second part | 4(iv), 8(b)(i), 10 first part |
| Basic vectors: parallel or collinear | — | 9(ii) | — |
| Basic vectors: dot product | — | 10(ii) | 6(i), 8(a), (b)(iii), 9(ii) |
| Basic vectors: cross product | — | 8(b) | — |
| Basic vectors: length of projection | — | 6 find | 2(ii), 3(ii), 4(ii), 5(ii), 6(ii) |
| Basic vectors: triangle or parallelogram area | — | 9(iii) | 3 last part, 5(iv), 6(iii), 8(b)(ii), 9(iii) |
Read the angle reference carefully. The printed table includes Example 27(ii) under angles between lines, but that ask is the internal angle OAB (130°), not the acute angle between the supporting lines. Its method uses the signed dot product. Example 26 and the linked acute-angle questions use the smaller line angle.
Review 4 answer-list slip. The closing lecture answer list prints under 4(iv). Those are the foot N’s coordinates from 4(iii). The reflection asked in 4(iv) is , as shown in the linked working. Keep N and D distinct.
Discussion 8 carries the printed reference [2017&2020/I/(modified)]. Its geometric interpretation must include the zero-direction cases. In Review 9, the closing answer list labels both the area and the final angle “(iii)”; the final angle belongs to ask (iv). Review 11’s printed “No” collision answer assumes the drone parameter is time; the linked solution explains the missing speed/time assumption.
Practice questions
33 examples and practice questions, in 6 sections.
More practice
Check every coordinate, state the parameter domain and interpret the result in the model. A supporting line, forward ray and time-dependent path can impose different conditions.
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