Multivariable-Calculus · Unit 4 · Video 1 · Interactive Practice

One Dot Product Writes the Plane: A Point, a Normal Vector, and Where d Comes From

IKey Formulas

Formula	Name	What you need
$\overrightarrow{P_0P}\cdot\mathbf{N} = 0$	Point-normal condition	One point $P_0$ of the plane and a normal $\mathbf{N}$
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$	Point-normal form, written out	$P_0 = (x_0, y_0, z_0)$ and $\mathbf{N} = \langle a, b, c\rangle$
$ax + by + cz = d$	Scalar equation of a plane	The coefficients are the components of $\mathbf{N}$
$d = ax_0 + by_0 + cz_0$	The constant, without expanding	The left-hand side evaluated at any point of the plane

Key Insight: $\mathbf{N}$ fixes the whole left-hand side, so parallel planes differ in one number only — and that number, $d$ , is the common value of $ax + by + cz$ at every point of the plane.

IIWhich Arrow Has to Be Perpendicular?

The dot product test needs a vector lying flat in the plane, so its tail must lie there too.

Plane

IIIFrom One Dot Product to $ax + by + cz = d$

Expanding the dot product moves every number of the point into a single constant.

Step 1

Step 1 — The vector from $P_0$ to $P$

\overrightarrow{P_0P} = \langle x - 2,\ y - 1,\ z - (-1)\rangle

\overrightarrow{P_0P} = \langle x - 2,\ y - 1,\ z + 1\rangle

Subtracting

-1

adds

1

: the third component is

z + 1

Step 2 — Perpendicular means dot product zero

\overrightarrow{P_0P}\cdot\mathbf{N} = 0, \qquad \mathbf{N} = \langle 1, 5, 10\rangle

1(x - 2) + 5(y - 1) + 10(z + 1) = 0

Already an equation of the plane, with the point and the normal both visible.

Step 3 — Expand the brackets

x - 2 + 5y - 5 + 10z + 10 = 0

Step 4 — Collect the constants

-2 - 5 + 10 = +3

x + 5y + 10z + 3 = 0

x + 5y + 10z = -3 \quad\checkmark

Step 5 — Check at $P_0$ , and read off the shortcut

2 + 5(1) + 10(-1) = 2 + 5 - 10 = -3 \quad\checkmark

The left-hand side evaluated at a point of the plane is

d

, so

d = -3

without expanding anything.

IVOne Normal Vector, a Whole Family

With $\mathbf{N}$ fixed only $d$ can change, and $d$ slides the plane along $\mathbf{N}$ .

d −3

💡 Multiplying the equation by $2$ gives $2x + 10y + 20z = -6$ : the constant doubles while the plane stays exactly where it was, so $d$ measures the slide only relative to $|\mathbf{N}| = \sqrt{126}$ .

VQuiz Questions

Problem 1 · A Point and a Normal

Given: the plane through $P_0 = (3, -1, 2)$ with normal vector $\mathbf{N} = \langle 2, 4, 1\rangle$ — find its equation.

✅ Correct! The components of

\mathbf{N}

are the coefficients, and

d = 2(3) + 4(-1) + 1(2) = 4

❌ That is the parallel plane through the origin.

d = 0

only when

P_0

itself satisfies

2x + 4y + z = 0

, and here it gives

4

❌ The roles are swapped.

\langle 3, -1, 2\rangle

are the coordinates of the point; the coefficients come from

\mathbf{N}

❌ Not quite. Evaluate

2x + 4y + z

P_0

and keep the sign it gives you.

Show solution

The components of the normal vector become the coefficients, so the left-hand side is $2x + 4y + z$ and only the constant is unknown:

2x + 4y + z = d

$P_0$ lies in the plane, so its coordinates satisfy the equation. Evaluate the left-hand side there:

d = 2(3) + 4(-1) + 1(2) = 6 - 4 + 2 = 4

\boxed{2x + 4y + z = 4}

The same answer by expanding the point-normal form: $2(x - 3) + 4(y + 1) + 1(z - 2) = 0$ gives $2x + 4y + z - 6 + 4 - 2 = 0$ , that is $2x + 4y + z = 4$ .

Problem 2 · Negative Coordinates in the Brackets

Given: the plane through $P_0 = (1, -2, -3)$ with normal vector $\mathbf{N} = \langle 4, 1, 5\rangle$ — find its equation.

✅ Correct!

4(1) + 1(-2) + 5(-3) = 4 - 2 - 15 = -13

❌ Check the brackets. With

y_0 = -2

the factor is

y - (-2) = y + 2

, not

y - 2

; likewise

z + 3

❌ Two slips at once. Writing

y - 2

and

z - 3

leads to

4x + y + 5z = 21

, not

-21

; with the brackets

y + 2

and

z + 3

the constant is

d = 4 - 2 - 15 = -13

❌ Not quite. The constant is

4x + y + 5z

evaluated at

P_0

, signs included.

Show solution

Point-normal form, watching both negative coordinates:

4\big(x - 1\big) + 1\big(y - (-2)\big) + 5\big(z - (-3)\big) = 0

4(x - 1) + (y + 2) + 5(z + 3) = 0

Expand and collect the constants, $-4 + 2 + 15 = 13$ :

4x + y + 5z + 13 = 0 \quad\Longrightarrow\quad \boxed{4x + y + 5z = -13}

Or take the shortcut: $d = 4(1) + 1(-2) + 5(-3) = 4 - 2 - 15 = -13$ .

Writing $(y - 2)$ and $(z - 3)$ instead would have produced $4x + y + 5z = 21$ — the plane through $(1, 2, 3)$ , a different plane altogether.

Problem 3 · Read the Normal, Then Slide the Plane

Given: the plane $2x - 3y + 6z = 12$ .

Which vector is normal to it?

Which $d$ makes $2x - 3y + 6z = d$ pass through $(1, 1, 1)$ ?

✅ Correct! Same normal

\langle 2, -3, 6\rangle

, new constant

2 - 3 + 6 = 5

: the plane

2x - 3y + 6z = 5

is parallel to the given one.

❌ Check the normal. The coefficients are read off in order, signs and all.

❌ Check the constant. Evaluate

2x - 3y + 6z

(1, 1, 1)

keeping the minus sign on the

y

term.

❌ That is the given plane's constant.

2x - 3y + 6z = 12

does not contain

(1, 1, 1)

: the left side there is

2 - 3 + 6 = 5

❌ $d = 0$ is the plane through the origin. The left side at

(1, 1, 1)

2 - 3 + 6 = 5

, not

0

Show solution

The normal. In $ax + by + cz = d$ the coefficient list is the normal vector, so

\mathbf{N} = \langle 2, -3, 6\rangle

The constant. A parallel plane keeps that normal, hence the same left-hand side; only $d$ changes. Since the new plane must contain $(1, 1, 1)$ , evaluate the left-hand side there:

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

\boxed{2x - 3y + 6z = 5}

Dropping the minus sign on the $y$ term gives $2 + 3 + 6 = 11$ , and keeping $d = 12$ returns the original plane, which does not contain $(1, 1, 1)$ since $5 \neq 12$ .

Problem 4 · Same Plane or a Different One?

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

✅ Correct! Both sides were multiplied by

2

: the normal becomes

\langle 2, 10, 20\rangle = 2\mathbf{N}

, still perpendicular to the same plane, and

P_0

still satisfies it.

❌ Only the left side was doubled. That equation is

x + 5y + 10z = -\tfrac{3}{2}

— parallel to the original, but not the same plane.

❌ The components were permuted.

\langle 10, 5, 1\rangle

points in a different direction, so it is the normal of a differently tilted plane.

❌ Not quite. Test

P_0 = (2, 1, -1)

in each equation: the same plane must still contain it.

Show solution

Two equations describe the same plane when one is a nonzero multiple of the other — both sides scaled together. Multiplying by $2$ :

2\big(x + 5y + 10z\big) = 2(-3) \quad\Longrightarrow\quad \boxed{2x + 10y + 20z = -6}

Check with the point: $2(2) + 10(1) + 20(-1) = 4 + 10 - 20 = -6\ \checkmark$ , and the new normal $\langle 2, 10, 20\rangle = 2\langle 1, 5, 10\rangle$ is the old one stretched, so it is perpendicular to the same plane.

The other options all fail that test at $P_0 = (2, 1, -1)$ :

$2x + 10y + 20z$ gives $-6 \neq -3$ ;
$x + 5y + 10z$ gives $-3 \neq 3$ ;
$10x + 5y + z$ gives $20 + 5 - 1 = 24 \neq -3$ .

Solved: 0 / 4

One Dot Product Writes the Plane: A Point, a Normal Vector, and Where d Comes From

IKey Formulas

IIWhich Arrow Has to Be Perpendicular?

IIIFrom One Dot Product to ax+by+cz=dax + by + cz = dax+by+cz=d

IVOne Normal Vector, a Whole Family

VQuiz Questions

IIIFrom One Dot Product to $ax + by + cz = d$