Vector/Cartesian equations of lines | Year 12

01 / LEARN

Understand the idea.

A line is a moving point

A spatial line has the form $r = a + t d$ , where $a$ locates one point and $d \neq 0$ gives direction. The same parameter must appear in all three components because they describe one moving point.

Two points determine the direction $d = b - a$ . To test another point, obtain $t$ from any nonconstant component, then check that identical value in the remaining components. Matching two coordinates alone does not establish membership.

Convert safely

When every direction component is nonzero, solve each component equation for $t$ and equate the results to obtain symmetric Cartesian form. A zero direction component instead gives a constant coordinate; it cannot appear as a denominator.

The whole line permits every real $t$ . The segment from $A$ to $B$ uses $a + t (b - a)$ with $0 \leq t \leq 1$ . A ray needs a one-sided interval, so always match the parameter range to the object requested.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Subtract the points to obtain a direction.
$d = (3, 1, 7) - (1, 2, 3) = (2, - 1, 4)$
Moving from $A$ to $B$ lies along the required line.
2
Anchor the line at A.
$r = (1, 2, 3) + t (2, - 1, 4), t \in ℝ$
Every real multiple of the nonzero direction traces the full line.
3
Eliminate the common parameter.
$\frac{x - 1}{2} = \frac{y - 2}{- 1} = \frac{z - 3}{4}$
All direction components are nonzero, so each component can be solved for the same $t$ .
4
Check the second endpoint.
$t = 1 ⟹ r = (3, 1, 7)$
The proposed line contains $B$ as well as its anchor $A$ .

Answer

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

EXAMPLE 02

Worked example 2

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

Worked solution

1
Read the component equations.
$x = 2, y = 1 + 2 t, z = - 3 + t$
The zero $x$ component in the direction makes $x$ constant.
2
Solve the changing coordinates for t.
$t = \frac{y - 1}{2}, t = z + 3$
Both changing coordinates describe the same parameter value.
3
State both Cartesian conditions.
$x = 2, \frac{y - 1}{2} = z + 3$
The constant coordinate and the remaining relation together specify the spatial line.

Answer

$x = 2$ , $(y - 1) / 2 = z + 3$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Yes.

The first coordinate gives $1 + 2 t = 5$ , hence $t = 2$ . The same value gives $y = 2 - 2 = 0$ and $z = 3 + 8 = 11$ , so the point is on the line.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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.
$x = 1$ , $\frac{y - 2}{4} = \frac{z - 3}{- 2}$ .
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 ↗

In this lesson

Understand the idea.

A line is a moving point

Convert safely

See the method in action.

Worked example 1

Subtract the points to obtain a direction.

Anchor the line at A.

Eliminate the common parameter.

Check the second endpoint.

Worked example 2

Read the component equations.

Solve the changing coordinates for t.

State both Cartesian conditions.

Try it yourself.

Check your understanding

Keep practising