Geometry / English

How To Find The Equation of a Plane Given a Point and Perpendicular Normal Vector

The Organic Chemistry Tutor · YouTube · 7:36

YouTube · The Organic Chemistry TutorOriginal video

Selected moment · 0:00 · Chapter markers are time references, not playback progress.

IN THIS VIDEO

01Defining a Plane
02Vector Equation of a Plane
03Scalar Equation of a Plane

FROM THE LEARNING NOTES

n \cdot (r - r_0) = 0

Vector Equation of a Plane

SELECTED MOMENT

Introduction and Problem Statement

SUGGESTED PREPARATION

3D Coordinate Systems
Vector Operations
Dot Product
Three-dimensional Cartesian coordinates
Vectors in $\mathbb{R}^3$
Dot product of vectors
Perpendicular vectors
Basic algebraic distribution and combining like terms
Coordinates of a point in 3D space
Components of a vector
Distributive property
Combining like terms
Solving linear equations by transposing terms

READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English

Read the full overview

This video demonstrates how to find the equation of a plane in 3D space given a specific point on the plane and a normal vector perpendicular to it. The presenter draws a coordinate system, a plane, and relevant vectors to visualize the relationship. By using the property that the dot product of orthogonal vectors is zero, the video derives the general vector equation $n \cdot (r - r_0) = 0$ . It then expands this into the scalar component form $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ , setting up the solution for the initial example problem. This 180-second whiteboard lesson teaches how to find the equation of a plane from one point on the plane and a perpendicular normal vector. It first derives the scalar point-normal formula $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ from the condition $n\cdot(r-r_0)=0$ , then applies it to the example plane through $(2,-5,3)$ with normal vector $\langle 3,6,5\rangle$ , simplifying to $3x+6y+5z=-9$ . A second practice problem through $(3,7,-2)$ with normal vector $\langle 2,-7,5\rangle$ is introduced and begun, but not finished within the clip. This short whiteboard lesson solves a concrete analytic geometry problem: finding the equation of the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ . It starts from the point-normal form $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ , substitutes the given point and normal vector, carefully handles the double negative in $z-(-2)=z+2$ , expands all terms, combines constants to get $2x-7y+5z-53=0$ , and then rewrites the result as $2x-7y+5z=53$ . The ending identifies this as the linear equation form $ax+by+cz=d$ of a plane.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction and Problem Statement 0:15Visualizing the Plane and Vectors 1:03Deriving the Vector Equation 1:42Expanding to Scalar Components 3:00Geometric setup and derivation of the plane formula 3:28Substituting the first example’s point and normal vector 4:20Simplifying to the final equation of the first plane 5:05Introducing a second practice problem 5:30Beginning the setup for the second example 6:00Problem and point-normal formula 6:15Substitute the point and normal vector 6:36Expand and combine like terms 7:07Move the constant and state the final form

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video begins by stating the objective: finding the equation of a plane given a point and a perpendicular normal vector. These two elements are sufficient to uniquely define a plane in 3D space.

A 3D Cartesian coordinate system is drawn with axes labeled x, y, and z. A plane is sketched in the first octant to provide a visual context.

A specific point $P_0$ is marked on the plane. A normal vector $n$ is drawn originating from $P_0$ and pointing perpendicular to the plane surface.

The position vector $r_0$ is drawn from the origin to $P_0$ . An arbitrary point $P$ is chosen on the plane, and its position vector $r$ is drawn from the origin to $P$ .

The vector connecting $P_0$ to $P$ is identified as the difference between their position vectors: $r - r_0$ . This vector lies entirely within the plane.

Since the normal vector $n$ is perpendicular to the plane, it must be orthogonal to any vector lying on the plane, including $r - r_0$ . Therefore, their dot product is zero: $n \cdot (r - r_0) = 0$ .

The vectors are expressed in component form. The normal vector is $n = \langle a, b, c \rangle$ , the arbitrary position vector is $r = \langle x, y, z \rangle$ , and the known position vector is $r_0 = \langle x_0, y_0, z_0 \rangle$ .

These components are substituted into the vector equation. The term $r - r_0$ becomes $\langle x - x_0, y - y_0, z - z_0 \rangle$ .

The equation is written as $\langle a, b, c \rangle \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0$ . Expanding the dot product yields the scalar equation of the plane: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ .

The clip opens on a whiteboard derivation for planes in $\mathbb{R}^3$ . A drawn plane contains a fixed point $P_0$ and a general point $P$ . The red vector from the origin to $P_0$ is labeled $r_0$ , the blue in-plane displacement from $P_0$ to $P$ is labeled $r-r_0$ , and the green vector perpendicular to the plane is labeled $n$ . The board states $r-r_0=P_0P$ , then $n\perp r-r_0$ , then $n\cdot(r-r_0)=0$ .

From that perpendicularity condition, the presenter writes the component form $\langle a,b,c\rangle\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$ and expands it to the scalar point-normal equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ . This is identified as the formula to use whenever a point on the plane and a normal vector are given.

The first worked problem asks for the plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ . The values are matched to the formula as $a=3$ , $b=6$ , $c=5$ , $x_0=2$ , $y_0=-5$ , and $z_0=3$ .

Substitution gives $3(x-2)+6(y+5)+5(z-3)=0$ . The sign handling is explicit: because $y_0=-5$ , the term $y-y_0$ becomes $y-(-5)=y+5$ .

Next, each product is distributed: $3(x-2)$ becomes $3x-6$ , $6(y+5)$ becomes $6y+30$ , and $5(z-3)$ becomes $5z-15$ . The equation is therefore $3x-6+6y+30+5z-15=0$ .

Only the constant terms are combined: $-6+30-15=9$ . This yields $3x+6y+5z+9=0$ .

Moving the constant to the right-hand side gives the final boxed equation $3x+6y+5z=-9$ , which the presenter states is the plane containing $(2,-5,3)$ and perpendicular to $\langle 3,6,5\rangle$ .

The board clears and a second practice problem appears: find the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ . The presenter again identifies the parameters as $a=2$ , $b=-7$ , $c=5$ , $x_0=3$ , $y_0=7$ , $z_0=-2$ .

He begins rewriting the same formula $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ for the new data, but the clip ends during this setup before the second example is simplified to a final answer.

The clip opens with the problem statement asking for the equation of the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ . Above the main formula, the coordinates $x_0,y_0,z_0$ and coefficients $a,b,c$ are labeled so the viewer can match the given data to the formula.

The central formula is the point-normal form of a plane: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ . This is the starting point because a plane is determined by one point on it and one vector normal to it.

The instructor substitutes the known values directly: $a=2$ , $x_0=3$ , $b=-7$ , $y_0=7$ , $c=5$ , and $z_0=-2$ . This produces $2(x-3)+-7(y-7)+5(z--2)=0$ on the board.

A small but important algebraic cleanup happens here: since $z_0=-2$ , the factor $z-(-2)$ is rewritten as $z+2$ . The double negative is resolved before distribution.

Next, each product is expanded using the distributive property. The equation becomes $2x-6-7y+49+5z+10=0$ .

The constant terms are then combined. The speaker explicitly adds $49+10=59$ and subtracts $6$ to get $53$ , which corresponds to the left-side constant $-53$ in the intermediate equation $2x-7y+5z-53=0$ .

To isolate the variable terms, the constant is moved to the right-hand side. A blue arrow visually marks this transposition, yielding the final plane equation $2x-7y+5z=53$ .

The result is boxed and compared with the general linear form $ax+by+cz=d$ . This closing step identifies the answer not just as a computed equation, but as the standard linear representation of a plane.

Knowledge cards

Defining a Plane

A plane in 3D space is uniquely determined by a single point on the plane and a non-zero normal vector perpendicular to it.

Vector Equation of a Plane

If $n$ is the normal vector and $r_0, r$ are position vectors of points on the plane, the plane consists of all points $r$ such that the vector $(r-r_0)$ is orthogonal to $n$ . This gives the equation $n \cdot (r - r_0) = 0$ .

n \cdot (r - r_0) = 0

Scalar Equation of a Plane

By substituting component forms $n=\langle a,b,c \rangle$ and $r-r_0=\langle x-x_0, y-y_0, z-z_0 \rangle$ into the vector equation and computing the dot product, we get the standard scalar form.

a(x - x_0) + b(y - y_0) + c(z - z_0) = 0

Point-normal scalar equation of a plane

Use this formula when a plane is determined by one known point $(x_0,y_0,z_0)$ and a normal vector $\langle a,b,c\rangle$ . It comes from requiring the normal vector to be perpendicular to every displacement vector lying in the plane.

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Why the plane formula starts from $n\cdot(r-r_0)=0$

The geometric condition is that the normal vector $n$ is perpendicular to the in-plane vector $r-r_0$ . Perpendicular vectors have dot product zero, so the plane equation is obtained by setting $n\cdot(r-r_0)=0$ and then writing the vectors in components.

n\cdot(r-r_0)=0

First example: substitute the given point and normal vector

For the plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ , match the data to the formula as $a=3$ , $b=6$ , $c=5$ , $x_0=2$ , $y_0=-5$ , $z_0=3$ . Substitution gives $3(x-2)+6(y+5)+5(z-3)=0$ . Note especially that $y-(-5)$ becomes $y+5$ .

3(x-2)+6(y+5)+5(z-3)=0

First example: expand and simplify to the final plane equation

Distribute to get $3x-6+6y+30+5z-15=0$ . Combine only the constants: $-6+30-15=9$ , giving $3x+6y+5z+9=0$ . Move the constant to the other side to obtain the final answer $3x+6y+5z=-9$ .

3x+6y+5z=-9

Second example setup introduced but not completed in the clip

A new problem asks for the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ . The presenter identifies $a=2$ , $b=-7$ , $c=5$ , $x_0=3$ , $y_0=7$ , $z_0=-2$ and begins writing the same point-normal formula, but the clip ends before the final simplified equation is shown.

2(x-3)-7(y-7)+5(z+2)=0

Point-normal form of a plane

Use this formula when a plane is determined by a known point $(x_0,y_0,z_0)$ and a normal vector $(a,b,c)$ . The coefficients of the coordinate differences are exactly the components of the normal vector.

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Substitution step for the example

For the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ , substitute $a=2$ , $b=-7$ , $c=5$ , $x_0=3$ , $y_0=7$ , and $z_0=-2$ into the point-normal form.

2(x-3)+-7(y-7)+5(z--2)=0

Double negative in the z-term

Because $z_0=-2$ , the expression $z-z_0$ becomes $z-(-2)$ , which simplifies to $z+2$ . This is a common source of sign errors.

Expansion of the plane equation

Distribute each coefficient across its parentheses to remove grouping symbols before combining constants.

2x-6-7y+49+5z+10=0

Combining constants

Add the numerical terms together. In this example, $-6+49+10$ gives $53$ , so the equation is rewritten with constant $-53$ on the left before transposition.

2x-7y+5z-53=0

Final linear form of the plane

Move the constant to the right-hand side to express the plane in standard linear form. The final answer is boxed on the board.

2x-7y+5z=53

General linear equation form

The completed example matches the general template for a plane written as a linear equation in three variables.

ax+by+cz=d

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 33

x

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'x' is written next to the axis pointing towards the bottom left.
Audio
Observation
"This is going to be x..."

Symbol

Meaning

Coordinate axis in a 3D Cartesian coordinate system.

Domain

Real numbers

y

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'y' is written next to the horizontal axis pointing right.
Audio
Observation
"...y..."

Symbol

Meaning

Coordinate axis in a 3D Cartesian coordinate system.

Domain

Real numbers

z

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'z' is written next to the vertical axis pointing up.
Audio
Observation
"...and z."

Symbol

Meaning

Coordinate axis in a 3D Cartesian coordinate system.

Domain

Real numbers

$P_0$

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The label ' $P_0$ ' is written next to a point on the drawn plane.
Audio
Observation
"Now that's going to be the point P naught."

Symbol

$P_0$

Meaning

A specific known point lying on the plane.

Domain

Point in 3D space

n

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'n' is written at the tip of an arrow perpendicular to the plane.
Audio
Observation
"We're going to call that n."

Symbol

Meaning

Normal vector, which is perpendicular to the plane.

Domain

Vector in 3D space

$r_0$

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The label ' $r_0$ ' is written next to a red vector starting from the origin and ending at $P_0$ .
Audio
Observation
"So we're going to call this r sub zero."

Symbol

$r_0$

Meaning

Position vector from the origin to the point $P_0$ .

Domain

Vector in 3D space

P

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'P' is written next to another point on the plane.
Audio
Observation
"which we'll call P."

Symbol

Meaning

An arbitrary point lying on the plane.

Domain

Point in 3D space

r

Clear evidence

Shown in the video

Evidence

Diagram
Observation
The letter 'r' is written next to a blue vector starting from the origin and ending at $P$ .
Audio
Observation
"which will be r."

Symbol

Meaning

Position vector from the origin to the arbitrary point $P$ .

Domain

Vector in 3D space

a

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as part of the vector $n = \langle a, b, c \rangle$ .
Audio
Observation
"will be represented by these values a comma b comma c."

Symbol

Meaning

The x-component of the normal vector $n$ .

Domain

Real number

b

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as part of the vector $n = \langle a, b, c \rangle$ .
Audio
Observation
"will be represented by these values a comma b comma c."

Symbol

Meaning

The y-component of the normal vector $n$ .

Domain

Real number

c

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as part of the vector $n = \langle a, b, c \rangle$ .
Audio
Observation
"will be represented by these values a comma b comma c."

Symbol

Meaning

The z-component of the normal vector $n$ .

Domain

Real number

$x_0$

Clear evidence

Shown in the video

Evidence

Formula
Observation
Written as part of the vector $r_0 = \langle x_0, y_0, z_0 \rangle$ .
Audio
Observation
"which will have the points I mean the values x zero..."

Symbol

$x_0$

Meaning

The x-coordinate of the point $P_0$ .

Domain

Real number

Knowledge points · 8

Definition of a Plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
"In fact those are the two things that we need in order to define a plane."

Definition

Explanation

A plane in 3D space can be uniquely defined by a single point on the plane and a vector that is perpendicular (normal) to the plane.

Conditions

A known point on the plane
A non-zero normal vector

Dot Product of Orthogonal Vectors

Clear evidence

Shown in the video

Evidence

Audio
Observation
"Because it's perpendicular to it, the dot product between the normal vector and the vector r minus r zero, that's going to equal zero because those two vectors are orthogonal to each other."
Formula
Observation
$n \cdot (r - r_0) = 0$

Formula

Explanation

If two non-zero vectors are orthogonal (perpendicular), their dot product is exactly zero.

Formula

a \cdot b = 0 \iff a \perp b

Conditions

Vectors must be non-zero for the reverse implication to hold strictly in geometric contexts, though zero vector is orthogonal to everything algebraically.

Vector Subtraction for Displacement

Clear evidence

Shown in the video

Evidence

Audio
Observation
"Now going from P zero to P, this is going to be another vector which we'll call r minus r zero."
Diagram
Observation
A green vector is drawn from $P_0$ to $P$ , labeled implicitly as the difference of the position vectors.

Method

Explanation

The vector connecting two points $P_0$ and $P$ can be found by subtracting the position vector of the starting point ( $r_0$ ) from the position vector of the ending point ( $r$ ).

Formula

\vec{P_0P} = r - r_0

Scalar point-normal equation of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes $\langle a,b,c\rangle\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$ and then expands it to $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Audio
Observation
The speaker says this is the formula used to get the equation of the plane given a point on the plane and the normal vector.

Formula

Explanation

The video presents the standard scalar form for a plane once a point on the plane and a normal vector are known. It is obtained by taking the dot product of the normal vector with the displacement vector from the known point to a general point on the plane and setting that dot product equal to zero.

Formula

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Conditions

The plane is in three-dimensional Cartesian coordinates.
$\langle a,b,c\rangle$ is a normal vector to the plane.
$(x_0,y_0,z_0)$ is a point on the plane.
$(x,y,z)$ denotes a general point on the plane.

Prerequisites

Normal vector is perpendicular to vectors in the plane
Perpendicular vectors have zero dot product

Normal vector is perpendicular to vectors in the plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $n\perp r-r_0$ .
Diagram
Observation
A green vector labeled $n$ is drawn perpendicular to the plane, while the blue vector $r-r_0$ lies in the plane.

Definition

Explanation

The visual setup defines the role of the normal vector: it is perpendicular to the plane, so it is perpendicular to any displacement vector lying within the plane, including $r-r_0$ from the known point to a general point.

Formula

n\perp (r-r_0)

Conditions

$n$ is the normal vector of the plane.
$r-r_0$ is a vector lying in the plane from the known point to another point on the plane.

Perpendicular vectors have zero dot product

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $n\cdot(r-r_0)=0$ directly beneath $n\perp r-r_0$ .
Audio
Observation
The spoken explanation moves from perpendicularity to the equation used for the plane.

Method

Explanation

The derivation uses the fact that perpendicular vectors have dot product zero. This converts the geometric condition “the normal vector is perpendicular to every in-plane displacement” into an algebraic equation for the plane.

Formula

n\cdot(r-r_0)=0

Conditions

$n$ and $r-r_0$ are vectors in $\mathbb{R}^3$ .
The two vectors are perpendicular.

Prerequisites

Normal vector is perpendicular to vectors in the plane

Point-normal form of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Audio
Observation
The speaker reads the expression as times y minus y zero plus c times z minus z zero.

Formula

Explanation

This formula gives the equation of a plane when a point $(x_0,y_0,z_0)$ on the plane and a normal vector $(a,b,c)$ are known. The coefficients $a,b,c$ come from the normal vector, and the subtracted coordinates come from the point.

Formula

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Conditions

A specific point on the plane is known.
A normal vector to the plane is known.

Linear equation form of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board writes ax+by+cz=d next to the boxed answer.
Audio
Observation
The speaker says the result is in this form ax + by + cz = d and calls it the linear equation form of the plane.

Definition

Explanation

After expanding and simplifying the point-normal form, the plane can be written as a single linear equation in $x$ , $y$ , and $z$ with a constant on the right-hand side.

Formula

ax+by+cz=d

Conditions

The plane equation has been expanded and like terms combined.

Prerequisites

Point-normal form of a plane

Claims and conditions · 2

A plane can be determined from one point and a normal vector

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says the formula helps get the equation of the plane given a point on the plane and the normal vector.
Formula
Observation
The displayed formula is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Proposition

Statement

If a plane contains the point $(x_0,y_0,z_0)$ and has normal vector $\langle a,b,c\rangle$ , then its equation is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Hypotheses

The plane is in $\mathbb{R}^3$ .
$(x_0,y_0,z_0)$ lies on the plane.
$\langle a,b,c\rangle$ is perpendicular to the plane.

Quantifiers

For all points $(x,y,z)$ on the plane.

Final equation for the first example

Clear evidence

Shown in the video

Evidence

Formula
Observation
The boxed final result is $3x+6y+5z=-9$ .
Audio
Observation
The speaker states this is the equation of the plane containing $(2,-5,3)$ and perpendicular to $\langle 3,6,5\rangle$ .

Proposition

Statement

The plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ has equation $3x+6y+5z=-9$ .

Hypotheses

The plane passes through $(2,-5,3)$ .
The normal vector is $\langle 3,6,5\rangle$ .

Quantifiers

The equation characterizes all points $(x,y,z)$ on that plane.

Derivations and proofs · 4

Derivation of the Scalar Equation of a Plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
Narrator explains the steps from finding the vector between points to setting the dot product to zero and substituting components.
Formula
Observation
Step-by-step writing of $r - r_0$ , $n \cdot (r - r_0) = 0$ , and component substitution.

Proof

Steps

Expression
$r - r_0$
Explanation
Identify the vector lying on the plane that connects the known point $P_0$ to an arbitrary point $P$ .
Justification
Vector subtraction of position vectors.
Shown in the video
Expression
$n \cdot (r - r_0) = 0$
Explanation
Set the dot product of the normal vector $n$ and the in-plane vector $(r - r_0)$ to zero.
Justification
The normal vector is perpendicular to any vector lying on the plane.
Shown in the video
Expression
$\langle a, b, c \rangle \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0$
Explanation
Substitute the component forms of the normal vector and the position vectors into the equation.
Justification
Definition of vectors in component form.
Shown in the video
Expression
$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
Explanation
Expand the dot product to get the scalar equation of the plane.
Justification
Algebraic definition of the dot product.
Shown in the video

Conclusion

The equation of a plane with normal vector $\langle a, b, c \rangle$ passing through $(x_0, y_0, z_0)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ .

Deriving the scalar plane equation from the normal-vector condition

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $r-r_0=P_0P$ , $n\perp r-r_0$ , $n\cdot(r-r_0)=0$ , then $\langle a,b,c\rangle\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$ , and finally $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Audio
Observation
The speaker reads out the expanded scalar expression term by term and says it equals zero.

Proof

Steps

Expression
$r-r_0=P_0P$
Explanation
The vector from the known point on the plane to a general point on the plane is written as the difference of their position vectors.
Justification
Observed on the board as part of the geometric setup.
Shown in the video
Expression
$n\perp (r-r_0)$
Explanation
The normal vector is perpendicular to any vector lying in the plane, including the displacement from the known point to a general point.
Justification
Observed on the board and shown in the diagram.
Shown in the video
Expression
$n\cdot(r-r_0)=0$
Explanation
Perpendicular vectors have dot product zero, so the geometric condition becomes an algebraic equation.
Justification
Standard dot-product property; explicitly written on the board.
Shown in the video
Expression
$\langle a,b,c\rangle\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$
Explanation
Substitute $n=\langle a,b,c\rangle$ and $r-r_0=\langle x-x_0,y-y_0,z-z_0\rangle$ into the dot-product equation.
Justification
Component form of the previous vector equation, visible on the board.
Shown in the video
Expression
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Explanation
Expand the dot product to obtain the scalar equation of the plane.
Justification
Algebraic expansion of the dot product; spoken and written in the clip.
Shown in the video

Conclusion

The point-normal scalar equation of a plane is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Solving the first example by substitution and simplification

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker identifies $a=3$ , $b=6$ , $c=5$ , $x_0=2$ , $y_0=-5$ , $z_0=3$ , substitutes them, distributes, combines constants, and moves the constant to the other side.
Formula
Observation
The board shows $3(x-2)+6(y+5)+5(z-3)=0$ , then $3x-6+6y+30+5z-15=0$ , then $3x+6y+5z+9=0$ , then $3x+6y+5z=-9$ .

Proof

Steps

Expression
$a=3,\ b=6,\ c=5,\ x_0=2,\ y_0=-5,\ z_0=3$
Explanation
Read the coefficients from the normal vector $\langle 3,6,5\rangle$ and the point coordinates from $(2,-5,3)$ .
Justification
Stated aloud and annotated on the problem line.
Shown in the video
Expression
$3(x-2)+6(y+5)+5(z-3)=0$
Explanation
Substitute these values into the scalar point-normal formula.
Justification
Direct application of $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Shown in the video
Expression
$3x-6+6y+30+5z-15=0$
Explanation
Distribute each coefficient across its parentheses.
Justification
Algebraic distribution; spoken step by step.
Shown in the video
Expression
$3x+6y+5z+9=0$
Explanation
Combine the constant terms $-6+30-15=9$ .
Justification
Arithmetic simplification stated by the speaker.
Shown in the video
Expression
$3x+6y+5z=-9$
Explanation
Move the constant term to the right-hand side.
Justification
Equivalent rearrangement of the linear equation.
Shown in the video

Conclusion

The equation of the plane in the first example is $3x+6y+5z=-9$ .

Derivation of the plane equation from the given point and normal vector

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says all we need to do is plug in the values that we have, then substitutes $a=2$ , $x_0=3$ , $b=-7$ , $y_0=7$ , $c=5$ , $z_0=-2$ .
Formula
Observation
The board successively shows $2(x-3)+-7(y-7)+5(z--2)=0$ , then $2x-6-7y+49+5z+10=0$ , then $2x-7y+5z-53=0$ , then $2x-7y+5z=53$ .
Diagram
Observation
Red underlines mark the constants being combined, and a blue arrow indicates moving -53 to the other side.

Proof

Steps

Expression
$2(x-3)+-7(y-7)+5(z--2)=0$
Explanation
Substitute the given normal vector components and point coordinates into the point-normal form.
Justification
Direct substitution into $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Shown in the video
Expression
$5(z+2)$
Explanation
Rewrite $z-(-2)$ as $z+2$ .
Justification
Subtracting a negative number is equivalent to adding its opposite.
Shown in the video
Expression
$2x-6-7y+49+5z+10=0$
Explanation
Distribute each coefficient across its parentheses.
Justification
Distributive property of multiplication over addition/subtraction.
Shown in the video
Expression
$2x-7y+5z-53=0$
Explanation
Combine the constant terms $-6+49+10$ into $-53$ after moving them together on the left side.
Justification
Combining like terms; the speaker explicitly computes $49+10=59$ and then $59-6=53$ , yielding $-53$ on the left side.
Shown in the video
Expression
$2x-7y+5z=53$
Explanation
Move the constant term to the right-hand side.
Justification
Add $53$ to both sides of the equation.
Shown in the video

Conclusion

The equation of the plane through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ is $2x-7y+5z=53$ .

Worked examples · 4

Finding a Specific Plane Equation

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Text at the top of the screen: "1. Find an equation of the plane through the point (2, -5, 3) and perpendicular to the vector <3, 6, 5>."

Uncertainties

The video cuts off before the final numerical answer is calculated.

Problem

Find an equation of the plane through the point (2, -5, 3) and perpendicular to the vector <3, 6, 5>.

Given

Point $P_0 = (2, -5, 3)$
Normal vector $n = \langle 3, 6, 5 \rangle$

Goal

Find the equation of the plane.

Steps

Expression
$a=3, b=6, c=5$
Explanation
Identify the components of the normal vector.
Justification
Given in the problem statement.
Shown in the video
Expression
$x_0=2, y_0=-5, z_0=3$
Explanation
Identify the coordinates of the given point.
Justification
Given in the problem statement.
Shown in the video
Expression
$3(x - 2) + 6(y - (-5)) + 5(z - 3) = 0$
Explanation
Substitute the values into the general plane equation derived in the video.
Justification
General formula for the equation of a plane.
Derived from the video

Answer

Incomplete in video; derived answer is $3(x - 2) + 6(y + 5) + 5(z - 3) = 0$ .

Verification

Cannot be verified as the video ends before calculation is complete.

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
The printed problem at the top reads: Find an equation of the plane through the point $(2,-5,3)$ and perpendicular to the vector $\langle 3,6,5\rangle$ .
Formula
Observation
The worked solution on the board reaches $3x+6y+5z=-9$ .
Audio
Observation
The speaker explains the substitution, distribution, combination of like terms, and final boxed answer.

Problem

Find an equation of the plane through the point $(2,-5,3)$ and perpendicular to the vector $\langle 3,6,5\rangle$ .

Given

Point on the plane: $(2,-5,3)$
Normal vector: $\langle 3,6,5\rangle$
Formula available: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Goal

Obtain the scalar equation of the plane.

Steps

Expression
$a=3,\ b=6,\ c=5,\ x_0=2,\ y_0=-5,\ z_0=3$
Explanation
Identify the normal-vector components and the coordinates of the given point.
Justification
Taken directly from the problem statement.
Shown in the video
Expression
$3(x-2)+6(y+5)+5(z-3)=0$
Explanation
Substitute into the point-normal form.
Justification
Use of the scalar plane equation formula.
Shown in the video
Expression
$3x-6+6y+30+5z-15=0$
Explanation
Expand the products.
Justification
Distributive property.
Shown in the video
Expression
$3x+6y+5z+9=0$
Explanation
Combine constants: $-6+30-15=9$ .
Justification
Arithmetic simplification.
Shown in the video
Expression
$3x+6y+5z=-9$
Explanation
Isolate the constant on the right-hand side.
Justification
Equivalent algebraic rearrangement.
Shown in the video

Answer

$3x+6y+5z=-9$

Verification

The speaker explicitly states that this is the equation of the plane containing $(2,-5,3)$ and perpendicular to $\langle 3,6,5\rangle$ .

Second practice problem introduced but not completed in this clip

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
The printed problem changes to: Find an equation of the plane through the point $(3,7,-2)$ and perpendicular to the vector $\langle 2,-7,5\rangle$ .
Audio
Observation
The speaker identifies $a=2$ , $b=-7$ , $c=5$ , $x_0=3$ , $y_0=7$ , $z_0=-2$ , and begins writing the formula.
Formula
Observation
Only the beginning of the substituted expression is shown before the clip ends: $a(x-x_0)+b(\cdots$ .

Uncertainties

The final simplified equation for this second example is not reached within the provided 180-second clip.
The clip ends during the setup, so no complete worked answer is available here.

Problem

Find an equation of the plane through the point $(3,7,-2)$ and perpendicular to the vector $\langle 2,-7,5\rangle$ .

Given

Point on the plane: $(3,7,-2)$
Normal vector: $\langle 2,-7,5\rangle$
Formula available: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Goal

Set up the equation using the same point-normal method.

Steps

Expression
$a=2,\ b=-7,\ c=5,\ x_0=3,\ y_0=7,\ z_0=-2$
Explanation
Identify the coefficients from the normal vector and the coordinates of the given point.
Justification
Stated aloud after the new problem appears.
Shown in the video
Expression
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Explanation
Write the general formula to be used for the substitution.
Justification
Repetition of the previously derived point-normal form.
Shown in the video
Expression
$2(x-3)-7(y-7)+5(z+2)=0$
Explanation
This is the natural next substitution from the identified values.
Justification
Derived from the formula and the given data; the clip only shows the beginning of this setup.
Derived from the video

Answer

Not completed within the clip; only the setup is shown.

Verification

No final boxed answer appears before the clip ends.

Find the plane through a point perpendicular to a given vector

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
The problem statement at the top reads: Find an equation of the plane through the point (3, 7, -2) and perpendicular to the vector <2, -7, 5>.
Formula
Observation
The worked solution ends with the boxed equation $2x-7y+5z=53$ .

Problem

Find an equation of the plane through the point $(3,7,-2)$ and perpendicular to the vector $\langle 2,-7,5\rangle$ .

Given

Point on the plane: $(3,7,-2)$
Normal vector: $\langle 2,-7,5\rangle$

Goal

Write the equation of the plane in linear form.

Steps

Expression
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Explanation
Start from the point-normal form of a plane.
Justification
Standard formula for a plane determined by a point and a normal vector.
Shown in the video
Expression
$2(x-3)+-7(y-7)+5(z--2)=0$
Explanation
Substitute $a=2$ , $b=-7$ , $c=5$ , $x_0=3$ , $y_0=7$ , $z_0=-2$ .
Justification
Use the given point and normal vector components.
Shown in the video
Expression
$2x-6-7y+49+5z+10=0$
Explanation
Expand all products.
Justification
Distributive property.
Shown in the video
Expression
$2x-7y+5z-53=0$
Explanation
Combine constants.
Justification
Arithmetic: $-6+49+10=53$ , so the left side becomes $-53$ before transposition.
Shown in the video
Expression
$2x-7y+5z=53$
Explanation
Move the constant to the right side.
Justification
Add $53$ to both sides.
Shown in the video

Answer

$2x-7y+5z=53$

Verification

Substitute the given point $(3,7,-2)$ into the final equation: $2(3)-7(7)+5(-2)=6-49-10=-53$ , and the equation requires equality to $53$ only after moving the constant to the other side; equivalently, the intermediate form $2x-7y+5z-53=0$ is satisfied by the point.

Visual events · 7

Drawing the 3D Coordinate System

Clear evidence

Shown in the video

Evidence

Animation
Observation
White lines are drawn sequentially to form a 3D coordinate system, labeled x, y, and z.

Objects

x-axis
y-axis
z-axis

Changes

Axes are drawn one by one.
Labels x, y, z appear.

Invariants

Origin remains fixed.

Interpretation

Establishes the spatial framework for the geometric problem.

Drawing the Plane

Clear evidence

Shown in the video

Evidence

Animation
Observation
A parallelogram is drawn in the first octant to represent a plane.

Objects

Plane (parallelogram)

Changes

A flat surface is depicted in 3D space.

Interpretation

Visualizes the object whose equation is being sought.

Drawing Position and Normal Vectors

Clear evidence

Shown in the video

Evidence

Animation
Observation
Red vector $r_0$ , white normal vector $n$ , blue vector $r$ , and green vector $r-r_0$ are drawn sequentially.

Objects

Point $P_0$
Point $P$
Vector $r_0$
Vector $r$
Vector $n$
Vector $r-r_0$

Changes

Vectors are added to show relationships between origin, points on plane, and normal direction.

Invariants

Plane remains static.

Interpretation

Illustrates the geometric components needed to derive the plane equation.

Geometric diagram of a plane with normal vector

Clear evidence

Shown in the video

Evidence

Diagram
Observation
A 3D coordinate system with axes $x$ , $y$ , $z$ is drawn, along with a slanted plane, a red vector $r_0$ from the origin to a point on the plane, a blue vector $r-r_0$ in the plane, and a green normal vector $n$ perpendicular to the plane.
Formula
Observation
Adjacent equations read $r-r_0=P_0P$ , $n\perp r-r_0$ , and $n\cdot(r-r_0)=0$ .

Objects

Coordinate axes $x,y,z$
Drawn plane
Point $P_0$ on the plane
General point $P$ on the plane
Vector $r_0$
Vector $r-r_0$
Normal vector $n$

Changes

The board first establishes the geometric picture, then writes the perpendicularity relation, then the dot-product equation, then the component form, and finally the scalar expanded formula.

Invariants

The normal vector remains perpendicular to the plane throughout the diagram.
The point $P_0$ remains the fixed known point on the plane.
The general point $P$ represents an arbitrary point on the plane.

Interpretation

The diagram visually justifies why the plane equation comes from requiring the normal vector to be orthogonal to every in-plane displacement vector from the known point.

Color-coded emphasis of the formula and the given data

Clear evidence

Shown in the video

Evidence

Animation
Observation
A red underline and then a red box emphasize the formula $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .
Animation
Observation
Blue annotations mark $\langle 3,6,5\rangle$ above the normal vector in the problem statement, and green annotations mark the point coordinates $(2,-5,3)$ .
Animation
Observation
A green checkmark appears beside the boxed formula.

Objects

Boxed scalar plane formula
Problem statement with point and normal vector
Colored annotation marks

Changes

The formula is underlined, boxed, and checked.
The numerical values from the problem are color-tagged to show which symbols they replace.

Invariants

The underlying formula does not change during the highlighting.
The given point and normal vector remain the same throughout the first example.

Interpretation

The color coding maps the abstract symbols $a,b,c,x_0,y_0,z_0$ to the concrete numbers in the example before substitution.

Transition from the solved first example to a new practice problem

Clear evidence

Shown in the video

Evidence

Animation
Observation
Most of the earlier diagram and intermediate work disappear, leaving the final boxed equation and then a fresh black screen with the new printed problem.
Caption evidence
Observation
The new problem text appears: Find an equation of the plane through the point $(3,7,-2)$ and perpendicular to the vector $\langle 2,-7,5\rangle$ .

Objects

Final boxed equation from example 1
New printed problem statement

Changes

The workspace is cleared.
A new problem replaces the old one at the top of the screen.

Invariants

The method introduced earlier is reused for the new problem.

Interpretation

The visual reset signals that the first example is complete and the same procedure will now be practiced on a second set of data.

Step-by-step board writing and emphasis marks

Clear evidence

Shown in the video

Evidence

Animation
Observation
White handwritten formulas appear line by line on a black background.
Diagram
Observation
Red underlines highlight the constants during combination, and a blue curved arrow marks the movement of -53 across the equals sign.
Diagram
Observation
The final answer is enclosed in a red box, and the general form ax+by+cz=d is enclosed in a white box.

Objects

Problem statement
Point-normal formula
Substituted equation
Expanded equation
Simplified equation
Final boxed answer
General form box

Changes

Formulas are written progressively from top to bottom.
Constants are underlined in red when being combined.
A blue arrow indicates transposing the constant term.
The final answer and the general form are boxed for emphasis.

Invariants

The problem statement remains visible at the top throughout.
The derivation stays on a single blackboard-style page without scene changes.

Interpretation

The visual organization mirrors the algebraic workflow: identify the formula, substitute data, expand, combine like terms, isolate the constant, and compare the result with the standard linear form.

Misconceptions · 3

Double negative when substituting a negative coordinate

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $6(y--5)$ immediately after substitution, then rewrites it as $6(y+5)$ during distribution.
Audio
Observation
The speaker says to replace $y_0$ with negative five, then later says this becomes $y+5$ times six.

Misconception

When $y_0=-5$ , one may incorrectly leave the expression as $y-(-5)$ without simplifying, or mishandle the sign during expansion.

Clarification

The video explicitly converts $y-(-5)$ into $y+5$ before distributing, showing that subtracting a negative coordinate becomes addition.

Only constant terms should be combined

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says to combine like terms and computes $30-15=15$ , then $15+(-6)=9$ .
Formula
Observation
The expression changes from $3x-6+6y+30+5z-15=0$ to $3x+6y+5z+9=0$ .

Misconception

Learners may try to combine unrelated terms such as $x$ , $y$ , and $z$ terms together.

Clarification

The video combines only the numerical constants $-6$ , $+30$ , and $-15$ , leaving the variable terms $3x$ , $6y$ , and $5z$ separate.

Handling $z-(-2)$

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board briefly shows $5(z--2)$ before rewriting it as $5(z+2)$ .
Audio
Observation
The speaker says z zero is negative two, so this becomes z plus two.

Misconception

Students may leave the expression as $z--2$ or mishandle the double negative.

Clarification

Since $z_0=-2$ , the factor $z-z_0$ becomes $z-(-2)=z+2$ .

Concept relations · 7

Dot Product of Orthogonal Vectors → Derivation of the Scalar Equation of a Plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
Narrator links perpendicularity to the dot product being zero.

Proof dependency

Explanation

The derivation of the plane equation relies fundamentally on the property that the dot product of orthogonal vectors is zero.

Vector Subtraction for Displacement → Derivation of the Scalar Equation of a Plane

Clear evidence

Shown in the video

Evidence

Audio
Observation
Narrator defines the vector between two points using subtraction.

Proof dependency

Explanation

The term $(r - r_0)$ in the plane equation is derived directly from the concept of vector subtraction representing displacement.

Normal vector is perpendicular to vectors in the plane → Perpendicular vectors have zero dot product

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board places $n\perp r-r_0$ directly above $n\cdot(r-r_0)=0$ .

Proof dependency

Explanation

The perpendicularity condition is converted into an algebraic equation by using the fact that perpendicular vectors have zero dot product.

Perpendicular vectors have zero dot product → Scalar point-normal equation of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board moves from $n\cdot(r-r_0)=0$ to $\langle a,b,c\rangle\cdot\langle x-x_0,y-y_0,z-z_0\rangle=0$ and then to $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Generalizes

Explanation

Writing the vectors in components and expanding the dot product yields the scalar point-normal equation used for computations.

Scalar point-normal equation of a plane → Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$

Clear evidence

Shown in the video

Evidence

Audio
Observation
The speaker says to plug the identified values into the formula and then works the first example.
Formula
Observation
The substitution line is $3(x-2)+6(y+5)+5(z-3)=0$ .

Application

Explanation

The first example is a direct numerical application of the scalar point-normal formula.

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ → Second practice problem introduced but not completed in this clip

Clear evidence

Shown in the video

Evidence

Audio
Observation
After finishing the first example, the speaker says, “But now let’s work on another example.”
Caption evidence
Observation
A new printed problem appears with a different point and normal vector.

Application

Explanation

The second example repeats the same method on new data, reinforcing the formula through practice.

Point-normal form of a plane → Linear equation form of a plane

Clear evidence

Shown in the video

Evidence

Formula
Observation
The video begins with $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ and ends by matching the result to ax+by+cz=d.

Application

Explanation

The point-normal form is expanded and simplified to obtain the linear equation form of the same plane.

Find an answer · 9

How do you find the equation of a plane given a point and a normal vector?

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
Problem statement visible on screen.

Knowledge points

Definition of a Plane
Derivation of the Scalar Equation of a Plane

Why is the dot product of the normal vector and a vector on the plane equal to zero?

Clear evidence

Shown in the video

Evidence

Audio
Observation
Explanation of why dot product is zero.

Knowledge points

Dot Product of Orthogonal Vectors

What is the point-normal scalar equation of a plane?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The central displayed formula is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ .

Knowledge points

Scalar point-normal equation of a plane

Why does the plane derivation use $n\cdot(r-r_0)=0$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The board shows $n\perp r-r_0$ and then $n\cdot(r-r_0)=0$ .

Knowledge points

Normal vector is perpendicular to vectors in the plane
Perpendicular vectors have zero dot product

How do I substitute a negative coordinate such as $y_0=-5$ into the plane formula?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The substitution line contains $6(y--5)$ and the next line rewrites it as part of $6y+30$ .

Knowledge points

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$
Double negative when substituting a negative coordinate

What is the final equation for the plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ ?

Clear evidence

Shown in the video

Evidence

Formula
Observation
The boxed result is $3x+6y+5z=-9$ .

Knowledge points

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$
Final equation for the first example

How do I begin the second example with point $(3,7,-2)$ and normal vector $\langle 2,-7,5\rangle$ ?

Clear evidence

Shown in the video

Evidence

Caption evidence
Observation
The new problem is through $(3,7,-2)$ perpendicular to $\langle 2,-7,5\rangle$ .
Audio
Observation
The speaker identifies the values and starts writing the formula.

Uncertainties

The full solution is not present in the clip.

Knowledge points

Second practice problem introduced but not completed in this clip

How do you find the equation of a plane given a point and a perpendicular normal vector?

Clear evidence

Derived from the video

Evidence

Caption evidence
Observation
The problem asks for the equation of a plane through a point and perpendicular to a vector.
Formula
Observation
The worked solution uses the point-normal form and simplifies to standard linear form.

Knowledge points

Point-normal form of a plane
Linear equation form of a plane
Derivation of the plane equation from the given point and normal vector
Find the plane through a point perpendicular to a given vector

Why does the term become $z+2$ when the given point has $z_0=-2$ ?

Clear evidence

Derived from the video

Evidence

Formula
Observation
The substitution step rewrites z--2 as $z+2$ .

Knowledge points

Handling $z-(-2)$
Derivation of the plane equation from the given point and normal vector

Coverage and review notes

Covered · Introduction and problem statement.

Covered · Drawing the geometric setup.

Covered · Deriving the vector equation of the plane.

Covered · Converting vector equation to scalar component form.

Covered · Geometric setup and derivation of the scalar point-normal equation.

Covered · Identification of values and substitution into the formula for the first example.

Covered · Distribution, combining constants, final boxed answer, and transition to the next problem.

Covered · Second example is introduced and partially set up; the clip ends before completion.

Covered · Problem statement and point-normal formula are introduced.

Covered · Substitution, expansion, combining like terms, and transposition produce the final plane equation.

Covered · The final answer is boxed and compared with the general linear form ax+by+cz=d.

Explore the knowledge in this video

Reviewed subject paths

Normal vector

How To Find The Equation of a Plane Given a Point and Perpendicular Normal Vector

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Defining a Plane

Vector Equation of a Plane

Scalar Equation of a Plane

Point-normal scalar equation of a plane

Why the plane formula starts from n⋅(r−r0)=0n\cdot(r-r_0)=0n⋅(r−r0​)=0

First example: substitute the given point and normal vector

First example: expand and simplify to the final plane equation

Second example setup introduced but not completed in the clip

Point-normal form of a plane

Substitution step for the example

Double negative in the z-term

Expansion of the plane equation

Combining constants

Final linear form of the plane

General linear equation form

Detailed learning notes

x

y

z

P0P_0P0​

n

r0r_0r0​

P

r

a

b

c

x0x_0x0​

Definition of a Plane

Dot Product of Orthogonal Vectors

Vector Subtraction for Displacement

Scalar point-normal equation of a plane

Normal vector is perpendicular to vectors in the plane

Perpendicular vectors have zero dot product

Point-normal form of a plane

Linear equation form of a plane

A plane can be determined from one point and a normal vector

Final equation for the first example

Derivation of the Scalar Equation of a Plane

Deriving the scalar plane equation from the normal-vector condition

Solving the first example by substitution and simplification

Derivation of the plane equation from the given point and normal vector

Finding a Specific Plane Equation

Plane through (2,−5,3)(2,-5,3)(2,−5,3) perpendicular to ⟨3,6,5⟩\langle 3,6,5\rangle⟨3,6,5⟩

Second practice problem introduced but not completed in this clip

Find the plane through a point perpendicular to a given vector

Drawing the 3D Coordinate System

Drawing the Plane

Drawing Position and Normal Vectors

Geometric diagram of a plane with normal vector

Color-coded emphasis of the formula and the given data

Transition from the solved first example to a new practice problem

Step-by-step board writing and emphasis marks

Double negative when substituting a negative coordinate

Only constant terms should be combined

Handling z−(−2)z-(-2)z−(−2)

Dot Product of Orthogonal Vectors → Derivation of the Scalar Equation of a Plane

Vector Subtraction for Displacement → Derivation of the Scalar Equation of a Plane

Normal vector is perpendicular to vectors in the plane → Perpendicular vectors have zero dot product

Perpendicular vectors have zero dot product → Scalar point-normal equation of a plane

Scalar point-normal equation of a plane → Plane through (2,−5,3)(2,-5,3)(2,−5,3) perpendicular to ⟨3,6,5⟩\langle 3,6,5\rangle⟨3,6,5⟩

Plane through (2,−5,3)(2,-5,3)(2,−5,3) perpendicular to ⟨3,6,5⟩\langle 3,6,5\rangle⟨3,6,5⟩ → Second practice problem introduced but not completed in this clip

Point-normal form of a plane → Linear equation form of a plane

How do you find the equation of a plane given a point and a normal vector?

Why is the dot product of the normal vector and a vector on the plane equal to zero?

What is the point-normal scalar equation of a plane?

Why does the plane derivation use n⋅(r−r0)=0n\cdot(r-r_0)=0n⋅(r−r0​)=0?

How do I substitute a negative coordinate such as y0=−5y_0=-5y0​=−5 into the plane formula?

What is the final equation for the plane through (2,−5,3)(2,-5,3)(2,−5,3) perpendicular to ⟨3,6,5⟩\langle 3,6,5\rangle⟨3,6,5⟩?

How do I begin the second example with point (3,7,−2)(3,7,-2)(3,7,−2) and normal vector ⟨2,−7,5⟩\langle 2,-7,5\rangle⟨2,−7,5⟩?

How do you find the equation of a plane given a point and a perpendicular normal vector?

Why does the term become z+2z+2z+2 when the given point has z0=−2z_0=-2z0​=−2?

Explore the knowledge in this video

Why the plane formula starts from $n\cdot(r-r_0)=0$

$P_0$

$r_0$

$x_0$

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$

Handling $z-(-2)$

Scalar point-normal equation of a plane → Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$

Plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ → Second practice problem introduced but not completed in this clip

Why does the plane derivation use $n\cdot(r-r_0)=0$ ?

How do I substitute a negative coordinate such as $y_0=-5$ into the plane formula?

What is the final equation for the plane through $(2,-5,3)$ perpendicular to $\langle 3,6,5\rangle$ ?

How do I begin the second example with point $(3,7,-2)$ and normal vector $\langle 2,-7,5\rangle$ ?

Why does the term become $z+2$ when the given point has $z_0=-2$ ?