Vector/Cartesian equations of lines | Year 12

01 / LEARN

The idea, step by step

A line is a moving point

A line through position a with nonzero direction d has equation $r = a + t d$ . All three component equations use the same parameter. Any nonzero scalar multiple of d gives the same direction.

To test a point, find t from one nonzero-direction coordinate and check all remaining coordinates. A match in only two components is not enough to establish membership in a spatial line.

Convert safely

When direction components are all nonzero, eliminate t to obtain $(x - a_{x}) / d_{x} = (y - a_{y}) / d_{y} = (z - a_{z}) / d_{z}$ . If one component of d is zero, state that coordinate as constant instead of dividing by zero.

The segment from A to B is $r = a + t (b - a)$ , $0 \leq t \leq 1$ . Outside this interval the same equation continues along the full line. Check whether the problem asks for a line, ray or segment.

A common trap

Each coordinate equation may use an unrelated parameter.

Instead: One shared parameter links the coordinates of each point on a line.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Write the line through A(1,2,3), B(3,1,7).

Subtract A from B for the direction.
Its Cartesian form is $(x - 1) / 2 = (y - 2) / (- 1) = (z - 3) / 4$ .

r = (1, 2, 3) + t (2, - 1, 4)

Example 2

Worked example 2

Convert $(2, 1, - 3) + t (0, 2, 1)$ to Cartesian conditions.

The x coordinate is constant.
Eliminate t using y and z only.

x = 2

(y - 1) / 2 = z + 3

Check your understanding

Is (5,0,11) on $(1, 2, 3) + t (2, - 1, 4)$ ?

Show answer

Yes.

The common parameter t=2 gives every coordinate.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Write a line through (2,-1,4) parallel to (1,3,-2).
2 marks
Find the point at t=-2 on $(1, 0, 3) + t (2, - 1, 1)$ .
2 marks
Is (3,2,4) on $(1, 0, 1) + t (1, 1, 2)$ ?
2 marks
Convert $(1, 2, 3) + t (0, 4, - 2)$ to Cartesian form.
2 marks
Parametrise the segment from (0,1,2) to (4,-1,6).
2 marks
Lines use points (1,2,3) and (3,1,7), and directions (2,-1,4) and (-4,2,-8). Are they the same line?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Write a line through (2,-1,4) parallel to (1,3,-2).
2 marks
Use the point as initial position.
Use the stated direction vector.
$r = (2, - 1, 4) + t (1, 3, - 2)$
Find the point at t=-2 on $(1, 0, 3) + t (2, - 1, 1)$ .
2 marks
Multiply the direction by -2.
Add $(- 4, 2, - 2)$ to the initial point.
$(- 3, 2, 1)$
Is (3,2,4) on $(1, 0, 1) + t (1, 1, 2)$ ?
2 marks
The x and y components require t=2.
Then z=5, not 4.
No.
Convert $(1, 2, 3) + t (0, 4, - 2)$ to Cartesian form.
2 marks
Keep x constant.
Eliminate t from the other two components.
$x = 1$ , $(y - 2) / 4 = (z - 3) / (- 2)$ .
Parametrise the segment from (0,1,2) to (4,-1,6).
2 marks
Subtract endpoints.
Restrict t to the segment interval.
$(0, 1, 2) + t (4, - 2, 4)$ , $0 \leq t \leq 1$ .
Lines use points (1,2,3) and (3,1,7), and directions (2,-1,4) and (-4,2,-8). Are they the same line?
2 marks
The directions are parallel.
The second point lies on the first line at t=1.
Yes.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics vector algebra, vector/Cartesian equations and linear systems. Original explanations and tasks.

Official curriculum source ↗

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