HL · Planes 1 · Equations of a plane

Module 25 of 46Lesson 1 of 7 · HL · Planes 1 · Equations of a plane

Planes: equations, intersections, angles & distances · Essential understanding 1

Equations of a plane

IB AA AHL 3.17with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A plane is a flat sheet that goes on for ever. It can be described in three ways: a vector equation (a point plus two directions), a normal form (a point plus the vector at right angles to the sheet) and a Cartesian equation

ax+by+cz=d

. The skill is moving between them, and the key that unlocks everything is the normal vector.

By the end of this lesson you can:

write the vector equation $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ of a plane
use the normal form $\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}$ and the Cartesian form $ax+by+cz=d$
find a normal vector with the cross product, and read one from a Cartesian equation
convert between the three forms, and test whether a point is on a plane

🧑🏾 Imagine this…

Lay a sheet of cardboard flat on a table. Now push a pencil straight up from the cardboard so that it is at 90° to the sheet. The pencil shows the tilt of the sheet, whatever the sheet’s size. If you know one point on the sheet and which way the pencil points, the sheet is fixed. That pencil is the normal vector. Another way to fix the sheet: three points that are not in a line (a camera tripod stands firmly on three legs for the same reason).

Normal vector n

a vector at 90° to every line in the plane. Any non-zero multiple of n is also a normal.

Direction vectors b, c

two vectors lying in the plane which are not parallel to each other.

Position vector a

the position vector of one known point A on the plane.

Cartesian equation

ax+by+cz=d

. The numbers

a,b,c

are the components of a normal vector.

★ Three forms of the same plane

Vector form: $\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}$ (λ and μ are any real numbers; b and c are not parallel).
Normal form: $\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}$ . Here a is any point on the plane.
Cartesian form: with $\mathbf{n}=\begin{pmatrix}a\\b\\c\end{pmatrix}$ and $\mathbf{r}=\begin{pmatrix}x\\y\\z\end{pmatrix}$ , the normal form becomes $ax+by+cz=d$ , where $d=\mathbf{a}\cdot\mathbf{n}$ .
Finding n: $\mathbf{n}=\mathbf{b}\times\mathbf{c}$ (cross product of two directions in the plane).
Three points A, B, C: directions $\overrightarrow{AB}$ and $\overrightarrow{AC}$ , so $\mathbf{n}=\overrightarrow{AB}\times\overrightarrow{AC}$ .
The vector and normal forms of a plane are in the IB formula booklet. You still have to know how to use them.

⚠️ Marks are lost here

The normal is not a point. In $2x-y+4z=12$ , the vector (2, −1, 4) is a direction, not a point on the plane.
To find d, put in a point on the plane. Do not use the normal vector for that.
Direction vectors b and c must not be parallel. If c = kb you only have a line.
Many answers are correct: a different point or different direction vectors give a different-looking but equal plane. Check by testing points.
Cartesian equations such as $4x-5y+3z=10$ can be multiplied by any non-zero number and describe the same plane.

💡 A fast check

After you find a Cartesian equation, substitute every point you were given. All should give the same d. This catches almost every slip in the cross product.

🧪 Try it yourselfA slice through the plane ax + by + cz = d

A computer screen is flat, so here we look at the slice of the plane where z = 0. That slice is the line

ax+by=d

, drawn as

y=\dfrac{d-ax}{b}

. The plane of Example 1 is

2x-y+4z=12

, so a = 2, b = −1 and d = 12. Change a, b and d: the line always stays at 90° to the pair (a, b), and changing only d slides the plane without tilting it.

Worked examples

Example 1.1A point and a normal vector

A plane passes through

A(1,2,3)

and has normal vector

\mathbf{n}=\begin{pmatrix}2\\-1\\4\end{pmatrix}

. Find the equation of the plane in normal form and in Cartesian form.

🧠 Think it through

Use $\mathbf{r}\cdot\mathbf{n}=\mathbf{a}\cdot\mathbf{n}$ with $\mathbf{a}=\overrightarrow{OA}$ .
Work out the scalar product a·n: that number is d.
Expand r·n with r = (x, y, z).

✍️ By hand
  (M1)
Normal form: 
Cartesian form:   (A1)

✅ Answer

\mathbf{r}\cdot(2,-1,4)^T=12

, i.e.

2x-y+4z=12

Check: A gives 2(1) − 2 + 4(3) = 12 ✓. This is the plane whose z = 0 slice is drawn in the lab above.

Example 1.2A plane through three points

Find (a) a vector equation and (b) the Cartesian equation of the plane through

A(1,0,2)

B(3,1,1)

and

C(2,2,4)

🧠 Think it through

Make two direction vectors from A: AB and AC.
The vector equation is a + λAB + μAC.
The normal is AB × AC. Then d = a·n.

✍️ By hand
  (A1)
(a) 
(b) 
  (M1)A1
Plane:   (A1)
Check B:  ✓. Check C:  ✓.

🔢 On your TI-Nspire CX II

Cross product: the TI has no cross-product key in exam mode, so type the formula as one vector: [1*2-(-1)*2,(-1)*1-2*2,2*2-1*1]. Then 4*1+(-5)*0+3*2 (n·a typed as components) gives d.
n = (4, −5, 3), d = 10

✅ Answer (a) r = (1, 0, 2) + λ(2, 1, −1) + μ(1, 2, 2). (b) 4x − 5y + 3z = 10.

In Paper 1 you do the cross product by hand. The GDC is a check, not a substitute.

Example 1.3Vector form to Cartesian form

A plane has equation

\mathbf{r}=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix}+\mu\begin{pmatrix}0\\1\\3\end{pmatrix}

. Find its Cartesian equation.

🧠 Think it through

The two direction vectors are in the plane, so n = b × c.
Use the point (1, 0, 2) to find d.
Or: write x, y, z in terms of λ and μ and eliminate them.

✍️ By hand
  (M1)A1
  (A1)
Elimination check: . Then 
, so , i.e.  ✓.

✅ Answer

6x-3y+z=8

Elimination is slower but needs no cross product. Use whichever you trust more.

Example 1.4Cartesian form to vector form

Write the plane

2x+y-z=4

in the form

\mathbf{r}=\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}

🧠 Think it through

Find three easy points by setting two coordinates to 0.
Subtract to get two direction vectors.
Check each direction against the normal (2, 1, −1): the scalar product must be 0.

✍️ By hand
Points: 
  (each satisfies the equation)
 use ;
 use 
Check:  ✓    ✓

✅ Answer r = (2, 0, 0) + λ(−1, 2, 0) + μ(1, 0, 2).

Another route: let y = λ and z = μ, so

x = 2 - \dfrac{\lambda}{2} + \dfrac{\mu}{2}

. This gives r = (2, 0, 0) + λ(−½, 1, 0) + μ(½, 0, 1). Both are correct: they are the same plane.

Example 1.5A plane containing a line and a point

The line

L:\ \mathbf{r}=\begin{pmatrix}1\\2\\0\end{pmatrix}+t\begin{pmatrix}1\\1\\2\end{pmatrix}

lies in a plane

\Pi

. The point

P(0,0,3)

is also on

\Pi

. Find the Cartesian equation of

\Pi

and show that the point

Q(2,3,2)

lies on it.

🧠 Think it through

Direction 1: the direction of L. Direction 2: from a point on L to P.
n = cross product of these. Then d from a point of L.
Substitute Q.

✍️ By hand
Take  on L. Then   (A1)
  (M1)A1
, so   (A1)
Check P:  ✓. Point Q:  ✓ so Q lies on .  (R1)

✅ Answer

7x-5y-z=-3

; Q satisfies it (LHS = −3), so Q is on the plane.

Q is on the plane because it is on L: when t = 1, L gives (2, 3, 2).

Challenge questions

⭐ Challenge 1.6Perpendicular to a linePaper 1 style

A plane

\Pi

is perpendicular to the line

\mathbf{r}=\begin{pmatrix}0\\1\\1\end{pmatrix}+t\begin{pmatrix}3\\-2\\1\end{pmatrix}

and passes through

A(2,-1,3)

. Find the Cartesian equation of

\Pi

, and show that

B(1,-4,0)

lies on

\Pi

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

A line perpendicular to the plane has direction equal to the normal.
Then d = a·n.

✍️ By hand
The direction  is a normal: .
, so .
B:  ✓

✅ Answer 3x − 2y + z = 11; B satisfies it (3 + 8 + 0 = 11).

⭐ Challenge 1.7A triangular roof panelPaper 2 style

Three corners of a flat roof panel are

P(1,2,-1)

Q(4,0,2)

and

R(3,3,0)

(units in metres).

(a) Find $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ , and hence a normal to the panel.
(b) Find the Cartesian equation of the plane of the panel.
(c) Find the area of triangle PQR.
(d) The point $S(k,-1,2)$ lies in the same plane. Find k.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

PQ × PR gives n; its length is also twice the area.
d = p·n.
Put S into the plane equation.

✍️ By hand
(a) 
;
(b) . So , i.e. . (Check Q:  ✓; R:  ✓)
(c) Area 
(d) 

🔢 On your TI-Nspire CX II

Check the area: type the cross product [(-2)*1-3*1,3*2-3*1,3*1-(-2)*2], then 0.5*norm(ans).
📍 Where to find it on the TI-Nspire
norm( is in menu💻 click menu on keypad▸7: Matrix & Vector▸7: Norms▸1: Norm

✅ Answer (a) PQ = (3, −2, 3), PR = (2, 1, 1), n = (−5, 3, 7). (b) 5x − 3y − 7z = 6. (c) 4.55 m². (d) k = 3.4.

The same cross product gives both the normal and the area. That double use is typical in exams.

Next: Lesson 2 · A line and a plane: intersection and angle

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Planes 1 · Equations of a plane

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Equations of a plane

Worked examples

Challenge questions