Multivariable-Calculus · Unit 4 · Video 2 · Interactive Practice

One Plane, Infinitely Many Equations: Read the Normal Vector Off the Coefficients

IKey Formulas

Formula	Name	What you need
$ax + by + cz = d$	Scalar equation of a plane	$\mathbf{N} = \langle a, b, c\rangle$ is a normal vector
$kax + kby + kcz = kd, \quad k \neq 0$	Rescaled equation	The same plane, with normal $k\mathbf{N}$
$\mathbf{v} = k\mathbf{N}, \quad k \neq 0$	$\mathbf{v}$ is perpendicular to the plane	A normal vector $\mathbf{N}$
$\mathbf{v} \cdot \mathbf{N} = 0$	$\mathbf{v}$ is parallel to the plane	A normal vector $\mathbf{N}$

Key Insight: The left-hand side of a plane equation is $\mathbf{N} \cdot \langle x, y, z\rangle$ . The coefficients carry the direction, and the constant $d$ only says which of the parallel planes with that normal you have — $d = 0$ being the one through the origin.

IIOne Plane, Infinitely Many Equations

Multiplying the whole equation by $k \neq 0$ rescales the normal vector and leaves the plane exactly where it was.

Multiplier (k) 1

IIIParallel, Perpendicular, or Neither

Two tests settle which of the three a vector is: $\mathbf{v} = k\mathbf{N}$ says perpendicular, $\mathbf{v} \cdot \mathbf{N} = 0$ says parallel.

💡 Challenge: exactly seven nonzero whole-number heads in range make $\mathbf{v} \cdot \mathbf{N} = 0$ — find all seven.

IVThe Head of the Vector

Plugging the components of $\mathbf{v}$ into the left-hand side computes $\overrightarrow{OP} \cdot \mathbf{N}$ , a test against the parallel plane through the origin.

Step 1

Step 1 — The head of $\mathbf{v}$

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

Drawn from the origin,

\mathbf{v}

has its head at the point

(1, 2, -1)

Step 2 — Plug the head into the left-hand side

x + y + 3z \ \text{ at } (1, 2, -1)

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

Step 3 — Compare with the right-hand side

0 \neq 5

The head is not a point of

x + y + 3z = 5

; that plane misses the origin, and the head sits off it.

Step 4 — Whose equation has right-hand side $0$ ?

x + y + 3z = 0

Same normal

\mathbf{N}

, constant

0

: the parallel plane through the origin. The head

(1, 2, -1)

lies in it.

\checkmark

Step 5 — The same dot product

\overrightarrow{OP} \cdot \mathbf{N} = \langle 1, 2, -1\rangle \cdot \langle 1, 1, 3\rangle = 0

The left-hand side evaluated at a point is

\overrightarrow{OP} \cdot \mathbf{N}

, so the two computations are one computation:

\mathbf{v} \perp \mathbf{N}

, hence

\mathbf{v}

is parallel to the plane.

\checkmark

VQuiz Questions

Problem 1 · Read the Normal Off the Coefficients

Given: the plane $4x - y + 6z = 11$ — find the normal vector whose components are the coefficients of $x$ , $y$ and $z$ .

✅ Correct! The coefficients are already the components of a normal vector — nothing has to be computed.

❌ Check the middle sign. The

y

term is

-y

, so its coefficient is

-1

, not

+1

. A sign there points the arrow into a different direction entirely.

❌ Those are the intercepts.

\tfrac{11}{4}

-11

and

\tfrac{11}{6}

are where the plane crosses the axes — points of the plane, not a direction perpendicular to it.

❌ That is $-\mathbf{N}$ . It is a perfectly good normal vector (every nonzero multiple is), but the vector written in the coefficients themselves is

\langle 4, -1, 6\rangle

Show solution

Match the equation against the general form $ax + by + cz = d$ :

4x + (-1)y + 6z = 11 \quad\Longrightarrow\quad a = 4,\ b = -1,\ c = 6,\ d = 11

The normal vector is $\mathbf{N} = \langle a, b, c\rangle = \langle 4, -1, 6\rangle$ . The constant $d = 11$ plays no part in it — it only selects which of the planes perpendicular to $\mathbf{N}$ this one is.

Check: $(1, -1, 1)$ and $(4, 11, 1)$ both satisfy the equation, so $\langle 3, 12, 0\rangle$ lies in the plane, and $\langle 4, -1, 6\rangle \cdot \langle 3, 12, 0\rangle = 12 - 12 + 0 = 0$ $\checkmark$

Problem 2 · Same Plane or a Different One?

Given: the plane $x + 5y + 10z = -3$ — which equation describes the same plane?

✅ Correct! Every term was multiplied by

3

, and dividing by

3

takes it straight back — so the two equations have exactly the same solutions.

❌ Only the left-hand side was scaled. Dividing by

3

gives

x + 5y + 10z = -1

, a parallel plane with the same normal, not this one. The constant has to be multiplied too.

❌ The coefficients were permuted. That equation has normal

\langle 10, 5, 1\rangle

, which is not a multiple of

\langle 1, 5, 10\rangle

— the two planes are tilted differently.

❌ Same normal, different constant.

\langle 1, 5, 10 \rangle

is still the normal, so this plane is parallel to ours — but it sits on the other side of the origin. Check

(2, 1, -1)

: it gives

-3

, not

3

Show solution

Multiplying an equation by a nonzero constant $k$ is a reversible step — divide by $k$ to undo it — so the solution set never changes:

3(x + 5y + 10z) = 3(-3) \quad\Longleftrightarrow\quad 3x + 15y + 30z = -9

The new coefficients $\langle 3, 15, 30\rangle = 3\mathbf{N}$ are three times as long and point the same way, so they are still perpendicular to the plane.

Check with the video's point $P_0 = (2, 1, -1)$ :

3(2) + 15(1) + 30(-1) = 6 + 15 - 30 = -9 \quad\checkmark

Options b and d keep the normal direction but change the constant, giving parallel planes; option c changes the normal direction altogether.

Problem 3 · Run the Test

Given: the plane $2x - y + 2z = 9$ and the vector $\mathbf{v} = \langle 1, 4, 1\rangle$ — test $\mathbf{v}$ against the plane.

What is $\mathbf{v} \cdot \mathbf{N}$ ?

So what is $\mathbf{v}$ to the plane?

✅ Correct!

\mathbf{v} \cdot \mathbf{N} = 0

makes

\mathbf{v}

perpendicular to the normal, and a vector perpendicular to the normal runs along the plane.

❌ A sign slipped. The middle coefficient is

-1

, so the middle term is

(4)(-1) = -4

, not

+4

❌ The constant is not part of the dot product. Only the coefficients

\langle 2, -1, 2\rangle

form

\mathbf{N}

; the

9

on the right says which parallel plane this is.

❌ Check the last term. The

z

coefficient is

+2

, so the last term is

(1)(2) = +2

2 - 4 + 2

, not

2 - 4 - 2

❌ Not quite.

\mathbf{v} \cdot \mathbf{N} = (1)(2) + (4)(-1) + (1)(2)

❌ That is the swap to watch for.

\mathbf{v} \cdot \mathbf{N} = 0

means

\mathbf{v} \perp \mathbf{N}

. Perpendicular to the normal means lying along the plane; perpendicular to the plane would mean

\mathbf{v} = k\mathbf{N}

❌ One of the two tests did fire. A zero dot product with

\mathbf{N}

is exactly the parallel test — "neither" is the verdict only when

\mathbf{v} \cdot \mathbf{N} \neq 0

and

\mathbf{v}

is not a multiple of

\mathbf{N}

❌ That is the point-versus-vector trap. At

(1, 4, 1)

the left-hand side is

2 - 4 + 2 = 0

, and

0 \neq 9

. The head lies in the parallel plane

2x - y + 2z = 0

through the origin, not in this one.

❌ Re-read the two tests.

\mathbf{v} = k\mathbf{N}

means perpendicular to the plane;

\mathbf{v} \cdot \mathbf{N} = 0

means parallel to it.

Show solution

Step 1: Read the normal off the coefficients.

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

Step 2: Is $\mathbf{v}$ a multiple of $\mathbf{N}$ ? Matching first components would need $k = \tfrac{1}{2}$ , but then the second component would be $-\tfrac{1}{2}$ , and $\mathbf{v}$ has $4$ . So $\mathbf{v} \neq k\mathbf{N}$ : it is not perpendicular to the plane.

Step 3: Dot it with $\mathbf{N}$ .

\mathbf{v} \cdot \mathbf{N} = (1)(2) + (4)(-1) + (1)(2) = 2 - 4 + 2 = 0

The dot product vanishes, so $\mathbf{v} \perp \mathbf{N}$ , and a vector perpendicular to the normal is parallel to the plane.

The head is a separate question: at $(1, 4, 1)$ the left-hand side is $0$ , and $0 \neq 9$ , so the head is not in this plane — it lies in the parallel copy $2x - y + 2z = 0$ through the origin. Sliding $\mathbf{v}$ so its tail sits at a point of the plane, say $(4, -1, 0)$ , puts its head at $(5, 3, 1)$ , and $10 - 3 + 2 = 9$ $\checkmark$ — flat in the plane.

Problem 4 · What Does the Zero Mean?

Given: the plane $x + 3y - z = 4$ and $\mathbf{v} = \langle 2, 1, 5\rangle$ . Substituting the components of $\mathbf{v}$ into the left-hand side gives $2 + 3 - 5 = 0$ . What does that zero establish?

✅ Correct! The left-hand side at a point is

\overrightarrow{OP} \cdot \mathbf{N}

, so one zero settles both statements at once.

❌ Perpendicular needs proportionality.

\mathbf{v}

would have to be

k\langle 1, 3, -1\rangle

; matching the first component gives

k = 2

, but then the second would be

6

, not

1

. A zero dot product says the opposite:

\mathbf{v}

runs along the plane.

❌ The plane demands $4$ , not $0$ . The point

(2, 1, 5)

gives

0

, so it is not on

x + 3y - z = 4

— it is on the parallel plane through the origin.

❌ The computation is a dot product.

x + 3y - z

evaluated at

(2, 1, 5)

\langle 2, 1, 5\rangle \cdot \langle 1, 3, -1\rangle

, so a zero there is precisely the test

\mathbf{v} \cdot \mathbf{N} = 0

Show solution

Read the normal off the coefficients: $\mathbf{N} = \langle 1, 3, -1\rangle$ . The substitution is literally a dot product:

x + 3y - z \ \text{ at } (2, 1, 5) \;=\; \langle 2, 1, 5\rangle \cdot \langle 1, 3, -1\rangle = 2 + 3 - 5 = 0

Reading 1 — the vector. $\mathbf{v} \cdot \mathbf{N} = 0$ , so $\mathbf{v} \perp \mathbf{N}$ , and a vector perpendicular to the normal is parallel to the plane.

Reading 2 — the point. The value $0$ is the right-hand side of $x + 3y - z = 0$ , the plane with the same normal through the origin. So the head of $\mathbf{v}$ lies in that parallel copy — and $0 \neq 4$ confirms it does not lie in the given plane.

Both readings are the same arithmetic, which is exactly why the head landing in the parallel plane through the origin means the vector runs parallel to the original plane.

Not perpendicular: $\mathbf{v} = k\mathbf{N}$ would force $k = 2$ from the first component and $k = \tfrac{1}{3}$ from the second — impossible.

Solved: 0 / 4