Vector equations of lines and planes: VCE Specialist Mathematics Unit 3
“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”
A line through point with direction has vector equation , ; splitting into coordinates gives the parametric form. A plane through point with normal satisfies , which expands to the Cartesian form . To find where a line meets a plane, substitute the parametric coordinates into the plane equation and solve for . The acute angle between a line and a plane satisfies .
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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 is the position vector of a known point and is a direction vector along the line, then every point on the line has position vector
Writing , , gives the parametric form , , . Eliminating from these (when no component of is zero) gives the symmetric Cartesian form .
The equation of a plane
A plane is fixed by one point on it and a normal vector perpendicular to it. A point lies on the plane exactly when the displacement is perpendicular to :
In coordinates with this is the Cartesian equation
The coefficients of 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 ; solve it, then back-substitute to get the intersection point. If the coefficient of 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 and normal give the angle between line and plane via
The sine appears (not cosine) because the angle to the plane is the complement of the angle to its normal.
Where a line pierces a plane
Find the point where the line meets the plane .
Step 1: Write the parametric coordinates of the line
Read off the components from the vector equation. Every point on the line satisfies:
This expresses each coordinate as a function of the single parameter , so substituting into the plane equation will yield one equation in one unknown.
Step 2: Substitute into the plane equation and solve for
Replacing , and in with the parametric expressions:
Expanding and collecting like terms:
Step 3: Back-substitute to find the intersection point
With , the parametric equations give:
The intersection point is . Verify it lies on the plane: . Correct.
- Reading the normal wrongly
- For the normal is ? No: it is , the coefficients of . Include every coefficient.
- Using cosine for the line-plane angle
- The angle to the plane uses , since 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 entirely, the line is parallel to the plane; check whether it lies in the plane.
Examples in context
Example 1. The plane through with normal is .
Example 2. The line through and has direction , so .
Try this
Q1. Write the vector equation of the line through with direction . [2 marks]
- Cue. .
Q2. State a normal vector to the plane . [1 mark]
- Cue. .
Q3. Does the point lie on the plane ? [2 marks]
- Cue. , 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 marksA 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 markConsider 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.)