Maths with Lemon

Equations of Planes

IB Mathematics: Analysis and Approaches Higher Level · Lesson V.6 · AHL 3.17

Learn to describe planes in three dimensions using vector, scalar-product and Cartesian equations, and convert between their forms.

Lesson V.6 worksheet (PDF) · Vectors overview

1. What Defines a Plane?

Key ideas

  • Three non-collinear points define a unique plane.
  • One point and two non-parallel direction vectors also define a unique plane.
  • A point and a non-zero normal vector determine a plane.
  • Two points alone lie on infinitely many planes.

Geometry in 3D

\[\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c\]

The parameters \(\lambda\) and \(\mu\) vary independently, generating every point on the plane.

Check

Why must \(\mathbf b\) and \(\mathbf c\) be non-parallel? If they were parallel, the equation would describe a line rather than a plane.

2. Vector Equation of a Plane

Steps

  • Take the position vector of a known point.
  • Choose two non-parallel vectors lying in the plane.
  • Introduce independent parameters.
  • To test a point, solve all three coordinate equations using the same two parameters.

Worksheet example

Point \(A(3,1,4)\) and direction vectors
\[\mathbf d_1=\begin{pmatrix}-1\\1\\3\end{pmatrix},\qquad\mathbf d_2=\begin{pmatrix}2\\0\\5\end{pmatrix}\]
\[\boxed{\mathbf r=\begin{pmatrix}3\\1\\4\end{pmatrix}+\lambda\begin{pmatrix}-1\\1\\3\end{pmatrix}+\mu\begin{pmatrix}2\\0\\5\end{pmatrix}}\]

For \(B(3,-2,5)\), the \(y\)-coordinate requires \(\lambda=-3\). The \(x\)-coordinate then requires \(\mu=-3/2\), but the resulting \(z\)-coordinate is \(-25/2\), not \(5\). Thus \(B\) is not on the plane.

3. Scalar-Product Form

Normal vectors

  • A normal vector is perpendicular to every direction vector in the plane.
  • The displacement between any two points of the plane is perpendicular to the normal.
  • Planes with parallel normal vectors are parallel or identical.

Point–normal equation

\[(\mathbf r-\mathbf a)\cdot\mathbf n=0\]
\[\boxed{\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n}\]
Worksheet: normal \(\mathbf n=(3,-2,2)\), point \(A(4,8,-1)\)
\[\mathbf a\cdot\mathbf n=12-16-2=-6\]
\[\boxed{\mathbf r\cdot(3,-2,2)=-6}\]

4. Cartesian Equation of a Plane

What to remember

  • The coefficients of \(x,y,z\) form a normal vector.
  • Substitute any point on the plane to determine the constant.
  • A Cartesian equation describes an entire plane, not just one point.

Standard form

\[\boxed{ax+by+cz=d}\]
\[\mathbf n=\begin{pmatrix}a\b\c\end{pmatrix}\]
Convert the preceding example
\[\boxed{3x-2y+2z=-6}\]

Verify by substituting \(A(4,8,-1)\).

5. Vector to Cartesian: Cross Product

Conversion method

  • Read the two direction vectors from the vector equation.
  • Find their cross product to obtain a normal.
  • Use a point from the vector equation to calculate \(d\).
  • Write \(ax+by+cz=d\) and simplify.

Worked example from the worksheet

Convert to Cartesian form
\[\mathbf r=\begin{pmatrix}1\\-2\\5\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\3\end{pmatrix}+\mu\begin{pmatrix}2\\-3\\5\end{pmatrix}\]
\[\mathbf n=\begin{pmatrix}1\\1\\3\end{pmatrix}\times\begin{pmatrix}2\\-3\\5\end{pmatrix}=\begin{pmatrix}14\\1\\-5\end{pmatrix}\]
\[d=14(1)+(-2)-5(5)=-13\]
\[\boxed{14x+y-5z=-13}\]

6. Plane Through Three Points

Construction

  • Select one point \(A\) as the starting position.
  • Form the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
  • Write the vector equation.
  • Cross the direction vectors to find a normal and then a Cartesian equation.

Worksheet example

Through \(A(3,4,-2)\), \(B(1,-1,3)\), \(C(5,0,2)\)
\[\overrightarrow{AB}=(-2,-5,5),\quad\overrightarrow{AC}=(2,-4,4)\]
\[\boxed{\mathbf r=\begin{pmatrix}3\\4\\-2\end{pmatrix}+\lambda\begin{pmatrix}-2\\-5\\5\end{pmatrix}+\mu\begin{pmatrix}2\\-4\\4\end{pmatrix}}\]
\[\overrightarrow{AB}\times\overrightarrow{AC}=(0,18,18)\]
\[\boxed{y+z=2}\]

7. Plane Containing Intersecting Lines

Procedure

  • Solve the two line equations to locate their intersection.
  • Take the two line direction vectors as plane directions.
  • Use the intersection as the plane’s fixed point.
  • Cross the direction vectors to find the normal.

Key formulas

\[\mathbf r=\mathbf p+t\mathbf u,\qquad\mathbf r=\mathbf q+s\mathbf v\]
\[\boxed{\mathbf r=\mathbf r_0+\lambda\mathbf u+\mu\mathbf v}\]

Here \(\mathbf r_0\) is the common point of the two intersecting lines. If the directions are parallel, a different construction is required.

8. Distance from a Point to a Plane

Shortest distance

  • The shortest segment is perpendicular to the plane.
  • Use the normal-vector coefficients in the denominator.
  • Substitute the point into the left-hand side and subtract the plane constant.

Formula and worksheet example

\[\boxed{D=\frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}}\]
Distance from the origin to \(x-2y+3z=2\)
\[D=\frac{|0-2|}{\sqrt{1+4+9}}=\boxed{\frac{2}{\sqrt{14}}}\]

9. Interactive Plane Explorer

Explore a plane in 3D

  • Adjust the coefficients in \(ax+by+cz=d\).
  • Rotate the diagram to inspect the orientation.
  • The arrow shows the normal vector.
  • The rendered grid lies on the plane, with perspective projected onto the screen.

Visualise the normal and plane

10. HL Practice and Revision

From Lesson V.6

  • Write vector equations and test whether points belong to planes.
  • Find scalar-product and Cartesian forms from a normal and point.
  • Convert vector equations to Cartesian equations using cross products.
  • Determine planes from three non-collinear points.
  • Find a plane containing two intersecting lines or a line and an external point.
  • Calculate the shortest distance from the origin to a plane.

Common mistakes

  • Using parallel direction vectors to define a plane.
  • Confusing a direction vector with a normal vector.
  • Taking the dot product instead of the cross product to find a normal.
  • Forgetting to substitute a known point to calculate \(d\).
  • Mixing the two parameters when checking whether a point lies on a plane.

Open the complete worksheet

Worksheet video: visualising scalar and vector products

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