Planes (AA HL)
A plane is a flat surface that stretches infinitely in two directions, and in IB vector work you'll need to pin one down algebraically in more than one form. This page covers the vector and Cartesian equations of a plane, how to move between them, and where marks are typically lost. It's part of the broader Vectors, Lines & Planes topic.
24 questions on this sub-topic.
The two forms of a plane's equation
Covered under IB syllabus reference AHL3.17: vector equations of a plane \(\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c\) (where \(\mathbf b\) and \(\mathbf c\) are non-parallel vectors in the plane) and \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\) (where \(\mathbf n\) is a normal to the plane), together with the Cartesian equation \(ax+by+cz=d\).
Normal (scalar product) form
\(\mathbf r\cdot\mathbf n = \mathbf a\cdot\mathbf n\)
Use when you're given (or can find) a normal vector \(\mathbf n\) and a point \(\mathbf a\) on the plane - this is usually the fastest route to the Cartesian equation.
Cartesian equation
\(ax+by+cz=d\)
Expand the normal form \(\mathbf n\cdot(\mathbf r-\mathbf a)=0\) and collect terms - the coefficients \(a,b,c\) are exactly the components of the normal vector \(\mathbf n\).
Need the full syllabus wording, intersections of planes, or the GDC methods? See Vectors, Lines & Planes.
Worked examples
Find the Cartesian equation of the plane through \((2,1,-1)\) with normal \((3,-2,1).\) Write your answer in the form \(ax+by+cz=d.\)
Worked solution
\(3(x-2) - 2(y-1) + 1(z+1) = 0.\) M1 A1
\(3x - 2y + z - 3 = 0.\) M1 A1
Find the point on the line \(\mathbf r=(1,0,2)+t(2,1,-2)\) closest to the origin.
Worked solution
the foot satisfies \(\mathbf r\cdot\mathbf d = 0\): \((1+2t, t, 2-2t)\cdot(2,1,-2)\) M1 \(= 0.\) A1
\(2(1+2t) + t - 2(2-2t) = 9t - 2\) M1 \(= 0.\) A1 \(t = \tfrac29.\) A1
\(\left(\tfrac{13}{9}, \tfrac29, \tfrac{14}{9}\right).\) A1
Common mistakes
- Reading off the wrong normal vector. In \(ax+by+cz=d\), the normal is \((a,b,c)\) exactly as written - not the coordinates of a point on the plane, which is a separate quantity entirely.
- Confusing a point on the plane with the normal vector. The vector form \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\) needs both a specific point \(\mathbf a\) and the normal \(\mathbf n\) - substituting one where the other belongs gives a completely different plane.
- Forgetting a plane needs two direction vectors, not one. Unlike a line, \(\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c\) requires two non-parallel vectors \(\mathbf b\) and \(\mathbf c\) lying in the plane - a single direction vector only describes a line within it.
Ready to practise properly?
24 plane-equation questions, marked instantly like the real exam.
Quick answers
What is the vector equation of a plane?
\(\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c\), where \(\mathbf a\) is a point on the plane and \(\mathbf b,\mathbf c\) are two non-parallel vectors that lie in it. Equivalently, \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\), where \(\mathbf n\) is a vector normal to the plane.
How do I find the Cartesian equation of a plane?
Write \(\mathbf n\cdot(\mathbf r-\mathbf a)=0\) using the normal vector \(\mathbf n=(a,b,c)\) and a known point, then expand to get \(ax+by+cz=d.\)