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NotesMath AA HLTopic 3.14Points on a line
Back to Math AA HL Topics
3.14.22 min read

Points on a line

IB Mathematics: Analysis and Approaches • Unit 3

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Contents

  • Is a point on the line? Find a point for a given λ
  • Where a line meets an axis, and lines as moving objects
One λ must work for ALL coordinates: A point lies on the line only if a single value of λ reproduces every coordinate.

Strategy: pick the easiest row, solve for λ, then check that same λ in the other rows. If they all match, the point is on the line; if even one disagrees, it isn't.

Going the other way is easier: to find the point at a particular λ, just substitute that number into r = a + λd.

IB-style question — does the point lie on the line?

A line has equation r = (1, 2, 0) + λ(2, −1, 3).

Determine whether the point Q(5, 0, 6) lies on the line.

Step by step

  1. Set r equal to Q and read off the three coordinate equations.
  2. Solve the first row for λ.
  3. Check λ = 2 in the other two rows.

Final answer

All three rows agree at λ = 2, so Q(5, 0, 6) lies on the line.

IB-style question — find the point at a given λ

For the line r = (4, −1, 2) + λ(1, 3, −2), find the position of the point when λ = 3.

Step by step

  1. Substitute λ = 3 into each coordinate.
  2. Multiply out and add.

Final answer

The point is (7, 8, −4).

Meeting an axis = the other coordinates are 0: A point is on the x-axis when its y- and z-coordinates are both 0; on the y-axis when x = z = 0; on the z-axis when x = y = 0.

So to find where a line crosses an axis: set the relevant coordinate equations to 0, solve for λ, then substitute back to get the crossing point. (In 2D, a line meets the x-axis where y = 0 and the y-axis where x = 0.)

IB-style question — where does the line cross an axis?

A 2D line has equation r = (6, 4) + λ(−2, 1).

Find where it crosses the x-axis.

Step by step

  1. On the x-axis the y-coordinate is 0. Set the y-row to 0.
  2. Substitute λ = −4 into the x-row to get the crossing point.

Final answer

The line crosses the x-axis at (14, 0).

Paper-2 model: a line that moves: On Paper 2 the same equation often describes a moving object (a boat, a drone, a helicopter):

r = (start position) + t(velocity), where the parameter t is time.

Then the velocity vector is d, and the speed is its magnitude |d|. Direction questions (e.g. an angle of descent) come from the components of d using trigonometry.
For a line r = a + t·d modelling motion, the speed is the magnitude of the velocity (direction) vector.

IB-style question — speed and angle of descent

A drone moves so that its position (in km) at time t hours is r = (1, 3, 2) + t(4, 0, −3).

(a) Find its speed. (b) Find the angle its path makes below the horizontal.

Step by step

  1. (a) Speed = magnitude of the velocity (4, 0, −3).
  2. (b) The descent is the downward part (z drops by 3) against the horizontal travel. Horizontal speed = √(4² + 0²) = 4.
  3. Take the inverse tangent.

Final answer

(a) Speed = 5 km/h. (b) Angle of descent ≈ 36.9° below the horizontal.

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For r = (0, 4, −2) + λ(3, −1, 2), find the coordinates of the point when λ = −1. [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

3.1.1Distance & midpoint (3D)
3.1.2Volume & surface area
3.1.3Angles in 3D
3.1.4Solids in 3D coordinates
View all Math AA HL topics

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3.14.1Vector equation of a line
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Angle between two lines3.15.1

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