🔬 Grade 12 · Lesson 9 of 13
Laser Plane: Planes in Three Dimensions
A single point to pin it and a single arrow to say which way it faces — that is all it takes to lock a flat plane into place anywhere in space.
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The big idea: In three dimensions, one equation of the form ax + by + cz = d describes a flat plane. The secret is the <strong>normal vector</strong> (a, b, c): it points straight out of the plane and fixes its tilt completely. Give a plane one point it must pass through and one normal direction, and there is exactly one plane that fits. From that idea come the equation of a plane, the angle between planes, and the distance from a point to a plane.
What pins a plane down?
A line in 2D is fixed by a point and a slope. A plane in 3D is fixed by a point and a direction it faces. That facing direction is a vector that sticks straight out of the plane, perpendicular to every line drawn within it. We call it the normal vector.
Tilt the normal and the whole plane swings to stay square to it, like a wall turning to stay perpendicular to the beam of a laser pointer. Fix the normal and fix one point, and the plane is completely determined.
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Normal + point = plane
A point P₀ on the plane and a normal vector n = (a, b, c) determine one unique plane. Every vector lying in the plane is perpendicular to n, so its dot product with n is zero — that single condition is the equation of the plane.
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
Point-normal form: n = (a, b, c) is the normal, and (x₀, y₀, z₀) is a known point on the plane.
The general form
Multiply out the point-normal form and collect the constants on the right. You get the compact general form:
Here the coefficients are the components of the normal vector — you can read the plane's tilt straight off the equation — and the constant d equals n · P₀, the dot product of the normal with any point on the plane.
\[ ax + by + cz = d, \qquad d = \vec{n}\cdot\vec{P_0} \]
General form. The normal is (a, b, c); d locates how far the plane sits from the origin along the normal.
🎮 Laser Plane 3D LIVE
A plane is fixed by a point and a normal vector — tilt the normal and watch the plane turn.
📝 Worked example: Find the equation of the plane through (1, 2, 3) with normal vector n = (2, −1, 4).
- Start from point-normal form: \( 2(x-1) - 1(y-2) + 4(z-3) = 0 \).
- Expand: \( 2x - 2 - y + 2 + 4z - 12 = 0 \).
- Collect: \( 2x - y + 4z - 12 = 0 \).
- Check with d = n·P₀ = 2(1) + (−1)(2) + 4(3) = 2 − 2 + 12 = 12. \checkmark
✓ The plane is <strong>2x − y + 4z = 12</strong>.
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The normal does the geometry
Two planes are parallel when their normals are parallel (scalar multiples of each other), and perpendicular when their normals are perpendicular (dot product zero). The angle between two planes is just the angle between their normals — every question about how planes sit reduces to a question about their normal vectors.
\[ \text{dist} = \dfrac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} \]
The perpendicular distance from a point (x₁, y₁, z₁) to the plane ax + by + cz = d.
📝 Worked example: How far is the point (1, 1, 1) from the plane x + 2y + 2z = 6?
- Plug the point into ax + by + cz − d: \( 1 + 2(1) + 2(1) - 6 = 1 + 2 + 2 - 6 = -1 \).
- Take the absolute value: \( |-1| = 1 \).
- Divide by the length of the normal: \( \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \).
✓ The distance is <strong>1/3</strong> of a unit.
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Length of the normal matters
The distance formula only works when you divide by the length of the normal. Forgetting the denominator √(a²+b²+c²) is the most common error — it gives an answer that depends on how you happened to scale the equation, which cannot be a real distance. Also keep the plane written as ax + by + cz = d so you subtract the correct constant.
Check your understanding
1. What is a normal vector to the plane 3x − 2y + z = 7?
In ax + by + cz = d, the coefficients (a, b, c) are the components of the normal vector, so n = (3, −2, 1).
2. Which point lies on the plane 2x + y − z = 2?
Test each in 2x + y − z: (1,0,0) gives 2 ✓. The others give (0,0,1)→−1, (0,2,3)→−1, and (0,1,1)→0, so only (1, 0, 0) is on the plane.
3. Two planes are parallel exactly when their normal vectors are…
Parallel planes face the same direction, so their normals point along the same line — one is a scalar multiple of the other.
4. What is the distance from the origin to the plane 2x − y + 2z = 9?
Plug (0,0,0): |0 − 9| = 9. Divide by |n| = √(4+1+4) = 3. So the distance is 9/3 = 3.
5. The plane through (0, 0, 5) with normal (0, 0, 1) has equation…
Point-normal: 0(x) + 0(y) + 1(z − 5) = 0, which simplifies to z = 5 — a horizontal plane at height 5.
✅ Key takeaways
- A plane in 3D is fixed by one point on it and one normal vector pointing straight out of it.
- Point-normal form a(x−x₀) + b(y−y₀) + c(z−z₀) = 0 expands to the general form ax + by + cz = d.
- The coefficients (a, b, c) are the normal vector; you read a plane's tilt straight off its equation.
- The angle between two planes is the angle between their normals; parallel planes have parallel normals.
- Distance from a point to a plane = |ax₁ + by₁ + cz₁ − d| divided by the length of the normal √(a²+b²+c²).