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Vector equations of lines and planes: VCE Specialist Mathematics Unit 3

Syllabus dot point

“Vector equations of lines and planes in three dimensions, their parametric and Cartesian forms, the use of a direction vector for a line and a normal vector for a plane, and the determination of intersections and the angle between a line and a plane”

6 min read

Quick answer

A line through point a\mathbf{a} with direction d\mathbf{d} has vector equation r=a+td\mathbf{r} = \mathbf{a} + t\mathbf{d}, t∈Rt \in \mathbb{R}; splitting into coordinates gives the parametric form. A plane through point a\mathbf{a} with normal n\mathbf{n} satisfies (r−a)⋅n=0(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0, which expands to the Cartesian form n1x+n2y+n3z=dn_1 x + n_2 y + n_3 z = d. To find where a line meets a plane, substitute the parametric coordinates into the plane equation and solve for tt. The acute angle ϕ\phi between a line and a plane satisfies sin⁡ϕ=∣d⋅n∣∣d∣∣n∣\sin\phi = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.

Jump to a section
  1. What this dot point is asking
  2. The vector equation of a line
  3. The equation of a plane
  4. Intersections and angles
  5. Examples in context
  6. Try this
  7. Exam-style questions

What this dot point is asking

VCAA wants you to write the vector equation of a line from a point and a direction, and of a plane from a point and a normal, to convert between vector, parametric and Cartesian forms, and to find intersections of lines with planes and the angles between them. This builds directly on the dot and cross products.

The vector equation of a line

A line is fixed by one point on it and a direction. If a\mathbf{a} is the position vector of a known point and d\mathbf{d} is a direction vector along the line, then every point on the line has position vector

r=a+td,t∈R.\mathbf{r} = \mathbf{a} + t\mathbf{d}, \qquad t \in \mathbb{R}.

Writing r=(x,y,z)\mathbf{r} = (x, y, z), a=(a1,a2,a3)\mathbf{a} = (a_1, a_2, a_3), d=(d1,d2,d3)\mathbf{d} = (d_1, d_2, d_3) gives the parametric form x=a1+td1x = a_1 + t d_1, y=a2+td2y = a_2 + t d_2, z=a3+td3z = a_3 + t d_3. Eliminating tt from these (when no component of d\mathbf{d} is zero) gives the symmetric Cartesian form x−a1d1=y−a2d2=z−a3d3\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}.

The equation of a plane

A plane is fixed by one point on it and a normal vector n\mathbf{n} perpendicular to it. A point r\mathbf{r} lies on the plane exactly when the displacement r−a\mathbf{r} - \mathbf{a} is perpendicular to n\mathbf{n}:

(r−a)⋅n=0,equivalentlyr⋅n=a⋅n.(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0, \qquad \text{equivalently} \qquad \mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}.

In coordinates with n=(n1,n2,n3)\mathbf{n} = (n_1, n_2, n_3) this is the Cartesian equation

n1x+n2y+n3z=d,d=a⋅n.n_1 x + n_2 y + n_3 z = d, \qquad d = \mathbf{a} \cdot \mathbf{n}.

The coefficients of x,y,zx, y, z are the components of a normal vector, which is how you read a normal straight off a Cartesian plane equation.

Intersections and angles

Line meets plane. Substitute the parametric coordinates of the line into the plane's Cartesian equation. This gives one equation in tt; solve it, then back-substitute to get the intersection point. If the coefficient of tt is zero, the line is parallel to the plane (no intersection, or the line lies in the plane).

Angle between a line and a plane. The direction d\mathbf{d} and normal n\mathbf{n} give the angle ϕ\phi between line and plane via

sin⁡ϕ=∣d⋅n∣∣d∣ ∣n∣.\sin\phi = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}|\,|\mathbf{n}|}.

The sine appears (not cosine) because the angle to the plane is the complement of the angle to its normal.

Worked example

Where a line pierces a plane

Find the point where the line r=(1,2,0)+t(2,−1,3)\mathbf{r} = (1, 2, 0) + t(2, -1, 3) meets the plane x+2y+z=10x + 2y + z = 10.

Step 1: Write the parametric coordinates of the line

Read off the components from the vector equation. Every point on the line satisfies:

x=1+2t,y=2−t,z=0+3t.x = 1 + 2t, \qquad y = 2 - t, \qquad z = 0 + 3t.

This expresses each coordinate as a function of the single parameter tt, so substituting into the plane equation will yield one equation in one unknown.

Step 2: Substitute into the plane equation and solve for tt

Replacing xx, yy and zz in x+2y+z=10x + 2y + z = 10 with the parametric expressions:

(1+2t)+2(2−t)+(3t)=10.(1 + 2t) + 2(2 - t) + (3t) = 10.

Expanding and collecting like terms:

1+2t+4−2t+3t=10⇒5+3t=10⇒t=53.1 + 2t + 4 - 2t + 3t = 10 \quad\Rightarrow\quad 5 + 3t = 10 \quad\Rightarrow\quad t = \frac{5}{3}.

Step 3: Back-substitute to find the intersection point

With t=53t = \frac{5}{3}, the parametric equations give:

x=1+2⋅53=133,y=2−53=13,z=3⋅53=5.x = 1 + 2\cdot\tfrac{5}{3} = \tfrac{13}{3}, \quad y = 2 - \tfrac{5}{3} = \tfrac{1}{3}, \quad z = 3\cdot\tfrac{5}{3} = 5.

The intersection point is (133,13,5)\left(\frac{13}{3}, \frac{1}{3}, 5\right). Verify it lies on the plane: 133+2⋅13+5=153+5=5+5=10\frac{13}{3} + 2\cdot\frac{1}{3} + 5 = \frac{15}{3} + 5 = 5 + 5 = 10. Correct.

Common exam traps
Reading the normal wrongly
For x+2y+z=10x + 2y + z = 10 the normal is (1,2,0)(1, 2, 0)? No: it is (1,2,1)(1, 2, 1), the coefficients of x,y,zx, y, z. Include every coefficient.
Using cosine for the line-plane angle
The angle to the plane uses sin⁡ϕ\sin\phi, since n\mathbf{n} is perpendicular to the plane.
One direction, one point per line
Any point on the line and any scalar multiple of the direction works, so equivalent equations can look different.
Forgetting the parallel case
If substituting eliminates tt entirely, the line is parallel to the plane; check whether it lies in the plane.

Examples in context

Example 1. The plane through (0,0,0)(0, 0, 0) with normal (1,1,1)(1, 1, 1) is x+y+z=0x + y + z = 0.

Example 2. The line through (1,0,0)(1, 0, 0) and (1,0,2)(1, 0, 2) has direction (0,0,2)(0, 0, 2), so r=(1,0,0)+t(0,0,1)\mathbf{r} = (1, 0, 0) + t(0, 0, 1).

Try this

Q1. Write the vector equation of the line through (2,1,5)(2, 1, 5) with direction (1,0,−2)(1, 0, -2). [2 marks]

  • Cue. r=(2,1,5)+t(1,0,−2)\mathbf{r} = (2, 1, 5) + t(1, 0, -2).

Q2. State a normal vector to the plane 3x−y+4z=73x - y + 4z = 7. [1 mark]

  • Cue. (3,−1,4)(3, -1, 4).

Q3. Does the point (1,1,1)(1, 1, 1) lie on the plane 2x+y−z=22x + y - z = 2? [2 marks]

  • Cue. 2+1−1=22 + 1 - 1 = 2, yes.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style2 marks
A plane contains the point A(1, 3, -2), and AB = -2i - 5j + 6k and AD = -i - j + 2k are two vectors lying in the plane. Hence find the equation of the plane in Cartesian form.
Show worked answer →

A normal to the plane is perpendicular to both AB and AD, so take the cross product n = AB x AD.

With AB = (-2, -5, 6) and AD = (-1, -1, 2):
i component: (-5)(2) - (6)(-1) = -4
j component: -[ (-2)(2) - (6)(-1) ] = -2
k component: (-2)(-1) - (-5)(-1) = -3
So n = (-4, -2, -3).

The plane has equation n . (r - A) = 0. Using point A(1, 3, -2):
-4(x - 1) - 2(y - 3) - 3(z + 2) = 0
-4x + 4 - 2y + 6 - 3z - 6 = 0
-4x - 2y - 3z + 4 = 0.

Multiplying by -1 gives the Cartesian equation 4x + 2y + 3z = 4. (Check with A: 4 + 6 - 6 = 4.)

2025 VCAA-style1 mark
Consider three planes defined by the equations P1: 2x + 9z = 8, P2: 3x + 6y + 5z = 7 and P3: x + 9y - 3z = 7. Find the point of intersection of the three planes.
Show worked answer →

Solve the three equations simultaneously.

From P1, 2x + 9z = 8, so x = (8 - 9z) / 2.

Substitute into P3: (8 - 9z) / 2 + 9y - 3z = 7. This gives 9y = 3 + (15/2)z, so y = 1/3 + (5/6)z.

Substitute x and y into P2: 3 . (8 - 9z) / 2 + 6(1/3 + (5/6)z) + 5z = 7, which simplifies to 14 - (7/2)z = 7, so z = 2.

Back-substitute: x = (8 - 18) / 2 = -5 and y = 1/3 + (5/6)(2) = 2.

The point of intersection is (-5, 2, 2). (Check in P2: -15 + 12 + 10 = 7.)

Practise this

Sources & how we know this

ExamExplained