Multivariable-Calculus · Unit 3 · Video 2 · Interactive Practice

The Plane Through Three Points: Flat Box, Normal Vector, One Equation

IKey Formulas

Formula	Name	What you need
$\det\big(\overrightarrow{P_1P},\ \overrightarrow{P_1P_2},\ \overrightarrow{P_1P_3}\big) = 0$	Flat-box test	Three points of the plane
$\mathbf{N} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}$	Normal from two in-plane vectors	Two vectors lying in the plane
$\overrightarrow{P_1P} \cdot \mathbf{N} = 0$	Point-normal form	One point and a normal
$n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0$	Scalar equation of the plane	$P_1 = (x_0, y_0, z_0)$ and $\mathbf{N} = \langle n_1, n_2, n_3\rangle$

Key Insight: The triple product is the determinant, $\overrightarrow{P_1P}\cdot\big(\overrightarrow{P_1P_2}\times\overrightarrow{P_1P_3}\big) = \det\big(\overrightarrow{P_1P},\overrightarrow{P_1P_2},\overrightarrow{P_1P_3}\big)$ , so both methods return the same number at every point of space — and the plane is the set of points where that number is zero.

IIThe Flat Box

The box built on $\overrightarrow{P_1P}$ , $\overrightarrow{P_1P_2}$ , $\overrightarrow{P_1P_3}$ is squashed flat exactly when $P$ lies in the plane.

💡 Challenge: exactly five of the 64 whole-number points in this box give the value zero — find all five.

IIIOne Plane, Many Normals

Every nonzero multiple of $\mathbf{N}$ is again normal to the plane, and each one produces the same plane.

Multiplier (c) 1

IVTwo Methods, One Equation

Written out in coordinates, the determinant and the dot product agree term for term.

Step 1

Step 1 — The three vectors from $P_1$

\overrightarrow{P_1P} = \langle x-1,\ y-1,\ z-1\rangle

\begin{aligned}\overrightarrow{P_1P_2} &= \langle -1, -1, 1\rangle \\ \overrightarrow{P_1P_3} &= \langle 2, -1, 0\rangle\end{aligned}

Step 2 — Method 1: the box is flat

\begin{vmatrix} x-1 & y-1 & z-1 \\ -1 & -1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = 0

\begin{aligned} (x-1)&\big[(-1)(0) - (1)(-1)\big] \\ -\ (y-1)&\big[(-1)(0) - (1)(2)\big] \\ +\ (z-1)&\big[(-1)(-1) - (-1)(2)\big] = 0 \end{aligned}

(x-1)(1) - (y-1)(-2) + (z-1)(3) = 0

Step 3 — Method 2: build the normal

\mathbf{N} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}

\begin{aligned} n_1 &= (-1)(0) - (1)(-1) = 1 \\ n_2 &= (1)(2) - (-1)(0) = 2 \\ n_3 &= (-1)(-1) - (-1)(2) = 3 \end{aligned}

\mathbf{N} = \langle 1, 2, 3\rangle \quad\checkmark

Step 4 — Method 2: dot it with $\overrightarrow{P_1P}$

\langle x-1,\ y-1,\ z-1\rangle \cdot \langle 1, 2, 3\rangle = 0

(x-1) + 2(y-1) + 3(z-1) = 0

Step 5 — Term by term

-(y-1)(-2) = +2(y-1)

Both lines read

(x-1) + 2(y-1) + 3(z-1) = 0

, that is

x + 2y + 3z = 6 \quad\checkmark

VQuiz Questions

Problem 1 · Normal Vector from Three Points

Given: $P_1 = (1, 1, 1)$ , $P_2 = (0, 0, 2)$ , $P_3 = (3, 0, 1)$ — find $\mathbf{N} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}$ .

✅ Correct! Both

\overrightarrow{P_1P_2}

and

\overrightarrow{P_1P_3}

lie in the plane, so their cross product is perpendicular to it — that is exactly what a normal vector is.

❌ Check the middle component. The middle entry of a cross product is

a_3b_1 - a_1b_3

, with the minus folded in — not

a_1b_3 - a_3b_1

. You have the sign of

n_2

reversed.

❌ That is the other order.

\overrightarrow{P_1P_3} \times \overrightarrow{P_1P_2} = -\mathbf{N}

. It is still a perfectly good normal vector, but it is not the product that was asked for.

❌ Not quite. A cross product is not computed component by component; each entry mixes two components of each vector.

Show solution

First the two fixed vectors:

\overrightarrow{P_1P_2} = \langle 0-1,\ 0-1,\ 2-1\rangle = \langle -1, -1, 1\rangle, \qquad \overrightarrow{P_1P_3} = \langle 3-1,\ 0-1,\ 1-1\rangle = \langle 2, -1, 0\rangle

Now the cross product, component by component:

$n_1 = a_2b_3 - a_3b_2 = (-1)(0) - (1)(-1) = 1$
$n_2 = a_3b_1 - a_1b_3 = (1)(2) - (-1)(0) = 2$
$n_3 = a_1b_2 - a_2b_1 = (-1)(-1) - (-1)(2) = 1 + 2 = 3$

\mathbf{N} = \langle 1, 2, 3\rangle

Check: $\mathbf{N}\cdot\overrightarrow{P_1P_2} = -1 - 2 + 3 = 0$ and $\mathbf{N}\cdot\overrightarrow{P_1P_3} = 2 - 2 + 0 = 0$ , so $\mathbf{N}$ is perpendicular to both.

Problem 2 · Point and Normal to Equation

Given: a plane through $P_1 = (1, 1, 1)$ with normal vector $\mathbf{N} = \langle 1, 2, 3\rangle$ — find its equation.

✅ Correct! The normal supplies the three coefficients, and the given point fixes the constant.

❌ That plane is parallel to the right one. The coefficients are correct, but

x + 2y + 3z = 0

passes through the origin. The constant has to come from

P_1

1 + 2 + 3 = 6

, not

0

❌ Check the constant. You added the coordinates of

P_1

. The constant is

\mathbf{N}\cdot P_1 = (1)(1) + (2)(1) + (3)(1) = 6

, each coordinate weighted by its normal component.

❌ The components of $\mathbf{N}$ are missing. Point-normal form is

n_1(x-x_0) + n_2(y-y_0) + n_3(z-z_0) = 0

; you used

n_1 = n_2 = n_3 = 1

❌ Not quite. Start from

\overrightarrow{P_1P}\cdot\mathbf{N} = 0

and expand.

Show solution

A point $P = (x, y, z)$ lies in the plane exactly when $\overrightarrow{P_1P} \perp \mathbf{N}$ , i.e. when the dot product vanishes:

\overrightarrow{P_1P}\cdot\mathbf{N} = 1(x-1) + 2(y-1) + 3(z-1) = 0

x - 1 + 2y - 2 + 3z - 3 = 0

x + 2y + 3z = 6

Check: $P_1 = (1,1,1)$ gives $1 + 2 + 3 = 6\ \checkmark$ . The point $(0,0,2)$ gives $0 + 0 + 6 = 6\ \checkmark$ , and $(3,0,1)$ gives $3 + 0 + 3 = 6\ \checkmark$ .

Problem 3 · A New Triple of Points

Given: $Q_1 = (1, 0, 0)$ , $Q_2 = (0, 2, 0)$ , $Q_3 = (0, 0, 4)$ — find the normal vector and then the equation of the plane through all three.

What is $\overrightarrow{Q_1Q_2} \times \overrightarrow{Q_1Q_3}$ ?

What is the equation of the plane?

✅ Correct!

\mathbf{N} = \langle 8, 4, 2\rangle

gives

8x + 4y + 2z = 8

, and dividing by

2

leaves

4x + 2y + z = 4

— the same plane, since scaling an equation never changes which points satisfy it.

❌ Check the middle component. It is

a_3b_1 - a_1b_3 = (0)(-1) - (-1)(4) = 4

, not

-4

: the minus of the cofactor expansion is already folded into that bracket.

❌ That is $\overrightarrow{Q_1Q_3} \times \overrightarrow{Q_1Q_2}$ . Reversing the two factors reverses the vector. It is still normal to the plane, but it is not the product asked for.

❌ Check the cross product. With

\overrightarrow{Q_1Q_2} = \langle -1, 2, 0\rangle

and

\overrightarrow{Q_1Q_3} = \langle -1, 0, 4\rangle

, the first component is

a_2b_3 - a_3b_2 = (2)(4) - (0)(0) = 8

❌ That plane passes through the origin. The coefficients are right, but the constant must come from a point:

\mathbf{N}\cdot Q_1 = 8(1) + 4(0) + 2(0) = 8

, so

8x + 4y + 2z = 8

❌ The coefficients are reversed. The coefficients of

x

y

z

are the components of

\mathbf{N} = \langle 8, 4, 2\rangle

in that order — after dividing by

2

\langle 4, 2, 1\rangle

❌ Divide the whole equation. Going from

8x + 4y + 2z = 8

4x + 2y + z

halves the left side, so the right side must be halved too:

4

, not

8

❌ Not quite. Substitute a given point into your equation —

Q_2 = (0,2,0)

and

Q_3 = (0,0,4)

must both satisfy it.

Show solution

Step 1 — Two vectors in the plane.

\overrightarrow{Q_1Q_2} = \langle -1, 2, 0\rangle, \qquad \overrightarrow{Q_1Q_3} = \langle -1, 0, 4\rangle

Step 2 — Their cross product is a normal vector.

$n_1 = (2)(4) - (0)(0) = 8$
$n_2 = (0)(-1) - (-1)(4) = 4$
$n_3 = (-1)(0) - (2)(-1) = 2$

\mathbf{N} = \langle 8, 4, 2\rangle

Step 3 — Point-normal form at $Q_1 = (1,0,0)$ .

8(x-1) + 4(y-0) + 2(z-0) = 0 \quad\Longrightarrow\quad 8x + 4y + 2z = 8

Step 4 — Divide by $2$ .

4x + 2y + z = 4

Check all three points: $Q_1: 4 + 0 + 0 = 4\ \checkmark$ $Q_2: 0 + 4 + 0 = 4\ \checkmark$ $Q_3: 0 + 0 + 4 = 4\ \checkmark$

Problem 4 · When Three Points Fail

Given: $R_1 = (1, 1, 1)$ , $R_2 = (2, 3, 5)$ , $R_3 = (3, 5, 9)$ — find $\overrightarrow{R_1R_2} \times \overrightarrow{R_1R_3}$ .

✅ Correct!

\overrightarrow{R_1R_3} = 2\,\overrightarrow{R_1R_2}

, so the two vectors are parallel and span no parallelogram: the cross product is

\mathbf{0}

. The three points lie on one line, so they determine no single plane — and the flat-box test is then passed by every point of space.

❌ That is one of the two vectors, not their cross product. Compute the components and watch what happens: every bracket cancels.

❌ Not quite. Compare

\overrightarrow{R_1R_2} = \langle 1, 2, 4\rangle

with

\overrightarrow{R_1R_3} = \langle 2, 4, 8\rangle

before computing anything — one is a multiple of the other.

Show solution

The two vectors from $R_1$ are

\overrightarrow{R_1R_2} = \langle 1, 2, 4\rangle, \qquad \overrightarrow{R_1R_3} = \langle 2, 4, 8\rangle = 2\,\overrightarrow{R_1R_2}

Component by component:

$n_1 = (2)(8) - (4)(4) = 16 - 16 = 0$
$n_2 = (4)(2) - (1)(8) = 8 - 8 = 0$
$n_3 = (1)(4) - (2)(2) = 4 - 4 = 0$

\overrightarrow{R_1R_2} \times \overrightarrow{R_1R_3} = \langle 0, 0, 0\rangle

The cross product of two parallel vectors is the zero vector, which is why the construction requires three points not on one line. With $\mathbf{N} = \mathbf{0}$ the condition $\overrightarrow{R_1P}\cdot\mathbf{N} = 0$ degenerates into $0 = 0$ : it holds at every point of space and singles out no plane at all. The determinant says the same thing — the box on three coplanar edge vectors has zero volume no matter where $P$ is.

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