Multivariable-Calculus · Unit 4 · Video 5 · Interactive Practice

Flat Box, Coplanar Normals: What a Zero Determinant Does to Three Planes

IKey Formulas

Formula	Name	What you need
$\det A = \det(\vec{N}_1, \vec{N}_2, \vec{N}_3)$	The rows of $A$ are the normals	The three plane equations
$\det A = 0 \iff \vec{N}_1, \vec{N}_2, \vec{N}_3$ coplanar	Flat box: volume $0$	The three normal vectors
$\vec{X} = t\,(\vec{N}_1 \times \vec{N}_2)$	Line of solutions of $A\vec{X} = \vec{0}$	$\vec{N}_1 \times \vec{N}_2 \neq \vec{0}$
$\det A \neq 0:\ \vec{X} = A^{-1}\vec{B}$ , exactly one	The two cases for $A\vec{X} = \vec{B}$	$\det A$ , then elimination if it is $0$

Key Insight: $A$ is built from the normals alone, so $A\vec{X} = \vec{0}$ and $A\vec{X} = \vec{B}$ carry the same determinant. A zero determinant rules out "exactly one solution" for both of them, and only elimination — ending at $0 = 0$ or at $1 = 2$ — decides between infinitely many and none.

IIThe Flat Box

The box on $\vec{N}_1, \vec{N}_2, \vec{N}_3$ has volume $|\det A|$ , so it is flat exactly when the normals are coplanar.

💡 Eight whole-number tips in this corner flatten the box, and one of them is $\vec{N}_3 = \vec{0}$ — whose equation $0 = d_3$ describes no plane at all.

IIIThe Line of Solutions

A normal built inside the plane of $\vec{N}_1$ and $\vec{N}_2$ is perpendicular to $\vec{N}_1 \times \vec{N}_2$ for free.

α in N₃ = αN₁ + βN₂ 1 β 1 t in X = t(N₁ × N₂) 0.5

IVNone or Infinitely Many

Seen down the shared direction $\vec{N}_1 \times \vec{N}_2$ , the three planes become three lines.

d₃ in 3x + 4y = d₃ 9

💡 Every point of this picture stands for a whole vertical line in space, so when the three lines do meet, the system's solution is that line — never a single point.

VQuiz Questions

Problem 1 · Read the Determinant Off the Normals

Given: three planes with normals $\vec{N}_1 = \langle 1, 2, 3\rangle$ , $\vec{N}_2 = \langle 2, -1, 1\rangle$ , $\vec{N}_3 = \langle 5, 0, 5\rangle$ — find $\det A$ for the system whose rows are these normals.

✅ Correct!

\vec{N}_3 = \vec{N}_1 + 2\vec{N}_2

, so the third normal already lies in the plane of the first two and the box on them is flat.

❌ Close, but check the signs.

20

is what the expansion gives with a plus in the middle; the cofactor signs alternate

+\ -\ +

❌ Not quite. Expand along the first row: the three minors are

-5

5

and

5

Show solution

Expanding along the first row, with the minors taken from rows 2 and 3:

\det A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 5 & 0 & 5\end{vmatrix} = 1\begin{vmatrix}-1 & 1 \\ 0 & 5\end{vmatrix} - 2\begin{vmatrix}2 & 1 \\ 5 & 5\end{vmatrix} + 3\begin{vmatrix}2 & -1 \\ 5 & 0\end{vmatrix}

= 1(-5) - 2(5) + 3(5) = -5 - 10 + 15 = 0

The reason is visible in the vectors themselves: $\vec{N}_1 + 2\vec{N}_2 = \langle 1 + 4,\ 2 - 2,\ 3 + 2\rangle = \langle 5, 0, 5\rangle = \vec{N}_3$ . A third edge that is a combination of the other two adds no thickness, so the box is flat and its volume — the size of the determinant — is $0$ .

Problem 2 · One Solution Found

Given: a system $A\vec{X} = \vec{B}$ with $\det A = 0$ , and a vector $\vec{X}_0 = \langle 1, 1, 1\rangle$ that satisfies all three equations — how many solutions does the system have?

✅ Correct! With

\det A = 0

exactly one solution is impossible, so finding one forces infinitely many.

❌ Close, but that is the one case $\det A = 0$ rules out. A unique solution needs

A^{-1}

, which needs

\det A \neq 0

❌ Not quite. A solution is given, so "none" is out; and the two cases available when

\det A = 0

are none and infinitely many.

Show solution

$\det A = 0$ leaves exactly two possibilities for $A\vec{X} = \vec{B}$ : no solution, or infinitely many. Here a solution is handed to us, so the first is excluded.

The mechanism: the normals are coplanar, so $\vec{V} = \vec{N}_1 \times \vec{N}_2$ is perpendicular to all three of them and solves the homogeneous system. Then for every $t$ ,

A(\vec{X}_0 + t\vec{V}) = A\vec{X}_0 + t\,A\vec{V} = \vec{B} + \vec{0} = \vec{B}

so a whole line of solutions runs through $\vec{X}_0$ in the direction $\vec{V}$ .

Problem 3 · Solve the Homogeneous System

Given: $A\vec{X} = \vec{0}$ with normals $\vec{N}_1 = \langle 1, 2, -1\rangle$ , $\vec{N}_2 = \langle 2, -1, 1\rangle$ , $\vec{N}_3 = \langle 3, 1, 0\rangle$ — find a nonzero solution and then the complete solution set.

A nonzero solution?

The complete solution set?

✅ Correct!

\vec{N}_3 = \vec{N}_1 + \vec{N}_2

, so the three planes share the line through the origin along

\vec{N}_1 \times \vec{N}_2

❌ Check the signs in the cross product.

\langle 1, 3, 5\rangle \cdot \vec{N}_1 = 1 + 6 - 5 = 2 \neq 0

, so this vector is not in the first plane.

❌ Not quite. A solution must be perpendicular to all three normals, so test it against

\vec{N}_1

first: a normal is never perpendicular to itself.

❌ Two is too few. If

\vec{X}

solves

A\vec{X} = \vec{0}

then so does every multiple:

2\langle 1, -3, -5\rangle = \langle 2, -6, -10\rangle

satisfies all three equations as well. The solution set is the whole line, not just the pair

\pm\langle 1, -3, -5\rangle

❌ Not quite.

\vec{N}_1

and

\vec{N}_2

are not parallel, so the three planes meet in a line, not in a point or a plane.

Show solution

Step 1 — the determinant. $\vec{N}_1 + \vec{N}_2 = \langle 3, 1, 0\rangle = \vec{N}_3$ , so the normals are coplanar and $\det A = 0$ : the homogeneous system has infinitely many solutions.

Step 2 — build one. Take the cross product of the first two normals:

\vec{N}_1 \times \vec{N}_2 = \big\langle (2)(1) - (-1)(-1),\ (-1)(2) - (1)(1),\ (1)(-1) - (2)(2) \big\rangle = \langle 1, -3, -5\rangle

Step 3 — check all three equations.

\vec{N}_1 \cdot \vec{X} = 1 - 6 + 5 = 0, \qquad \vec{N}_2 \cdot \vec{X} = 2 + 3 - 5 = 0, \qquad \vec{N}_3 \cdot \vec{X} = 3 - 3 + 0 = 0

The third costs nothing: $\vec{N}_3$ lies in the plane of $\vec{N}_1$ and $\vec{N}_2$ , and $\vec{X}$ is perpendicular to that whole plane.

Step 4 — the full set. Every multiple $t\langle 1, -3, -5\rangle$ satisfies all three equations, and together they fill the line of intersection.

Problem 4 · Same $A$ , New Right-Hand Side

Given: $x + 2y - z = 3$ , $\ 2x - y + z = 1$ , $\ 3x + y = 7$ — find $\det A$ and the number of solutions.

What is $\det A$ ?

How many solutions?

✅ Correct! The same coefficients as Problem 3, so

\det A = 0

again — but this right-hand side ends elimination at a contradiction.

❌ That is the other ending. It needs

d_3 = 4

: adding the first two equations gives

3x + y = 4

, while the third demands

3x + y = 7

❌ Recount the expansion. The third row is the sum of the first two,

\langle 1, 2, -1\rangle + \langle 2, -1, 1\rangle = \langle 3, 1, 0\rangle

, so the box is flat:

1(0 - 1) - 2(0 - 3) + (-1)(2 + 3) = -1 + 6 - 5 = 0

❌ Not quite. The rows of

A

are

\langle 1, 2, -1\rangle

\langle 2, -1, 1\rangle

\langle 3, 1, 0\rangle

, and the right-hand side never enters

A

❌ Not quite. Add the first two equations and compare the result with the third.

Show solution

Step 1 — the determinant. The rows of $A$ are $\langle 1, 2, -1\rangle$ , $\langle 2, -1, 1\rangle$ and $\langle 3, 1, 0\rangle$ , and the third is the sum of the first two, so the normals are coplanar and $\det A = 0$ . The right-hand side $\langle 3, 1, 7\rangle$ plays no part in this: $A$ holds only the coefficients.

Step 2 — eliminate. Adding the first two equations:

(x + 2y - z) + (2x - y + z) = 3 + 1 \quad\Longrightarrow\quad 3x + y = 4

The third equation says $3x + y = 7$ . Subtracting gives

0 = 3,

a contradiction, so the system has no solutions: the three planes are all parallel to $\vec{N}_1 \times \vec{N}_2 = \langle 1, -3, -5\rangle$ but have no common point.

Step 3 — the other ending. Replace the third right-hand side by $4$ and elimination ends at $0 = 0$ : the solutions are then the whole line $\langle 1, 1, 0\rangle + t\langle 1, -3, -5\rangle$ . Same $A$ , same $\det A = 0$ , opposite verdict — which is exactly why the determinant alone cannot decide.

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