Solving Linear Systems and Plane Equations

It is forbidden to kill; therefore all murderers are punished unless they kill in large numbers and to the sound of trumpets, Voltaire

Recall: Vector Basics

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.

Basic Vector Operations

• Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.

• Vector Subtraction: The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.

• The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$
  Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$

• The projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.

• The cross product, denoted by $\vec{A} \times \vec{B}$, is a binary operation on two vectors in three-dimensional space. It yields a new vector that is perpendicular (orthogonal, at a right angle) to both original vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
  The cross product $\vec{A}x\vec{B}$ can be computed using the determinant of a $3 \times 3$ matrix formed by the standard basis vectors ($\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$) and the components of the two vectors: $\vec{A} \times \vec{B} = det(\begin{smallmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|\vec{k}$.
  Example: If $\vec{A} = \langle 1, 2, 3 \rangle$ and $\vec{B} = \langle 4, 5, 6 \rangle$: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = (2\cdot 6 - 3\cdot 5)\hat{\mathbf{i}} - (1\cdot 6 - 3\cdot 4)\hat{\mathbf{j}} + (1\cdot 5 - 2\cdot 4)\hat{\mathbf{k}} = -3\hat{\mathbf{i}} + 6\hat{\mathbf{j}} - 3\hat{\mathbf{k}} = \langle -3, 6, -3 \rangle$

• Matrices offers an efficient way for solving systems of linear equations. Utilizing matrices, we can represent these equations more concisely and conveniently.

  For a system like, $\Bigl(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}\Bigr) \Bigl(\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}u_1\\ u_2\\ u_3\end{smallmatrix}\Bigr)$, $A \cdot X = U$ provides a more compact notation and also an efficient way to solve system of linear equations. A is a $3 \times 3$ matrix representing the coefficients, X is a $3 \times 1$ column vector representing the variables and U is a $3 \times 1$ column vector representing the results.

• Given a square matrix A (a matrix that has an equal number of rows and columns), an inverse matrix $A^{-1}$ exists if and only if A is non-singular, meaning its determinant is non-zero, $det(A) \neq 0$.

  The property of an inverse matrix is: $A \times A^{-1} = A^{-1} \times A = I$ where I is the identity matrix, which is a matrix with 1’s on the diagonal and 0’s elsewhere. Essentially, multiplying a matrix by its inverse reverses (“undoes”) the effect of the original matrix.

Solving Systems of Linear Equations

Matrices offer an incredibly efficient way to solve systems of linear equations. By separating the variables from their coefficients, we can represent complex mathematical systems concisely and solve them using algorithmic steps —this is exactly how computers handle massive systems in physics, engineering, and data science.

Consider a generic system of linear equations:

$$\begin{cases} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\ \dots \\ a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n \end{cases}$$

This can be elegantly compressed into the matrix equation $AX = B$, where:

1. A is a $n \times n$ matrix known as the coefficient matrix ($a_{ij}$).
2. X is an $n \times 1$ column matrix (vector) of variables ($x_1, x_2, \dots, x_n$).
3. B is an $n \times 1$ column matrix (vector) of the constants on the right side of the equations ($b_1, b_2, \dots, b_n$).

The Solution Logic

Solving a system of linear equations expressed as AX = B involves finding or isolating the matrix X when we are given both matrices A and B. Because matrix division doesn’t technically exist, we use the inverse matrix instead.

To solve AX = B, follow these steps:

1. Ensure A is non-singular. This means that we need to check that the determinant of A is not zero ($\det(A) \neq 0$).
   If it is zero, the inverse doesn’t exist, and the system does not have a single unique solution.
2. Calculate the inverse of A.
3. Multiply both sides of the equation by $A^{-1}$: $A^{-1}(AX) = A^{-1}B$.
   Matrix multiplication is associative, so we can regroup the terms: $(A^{-1}A)X = A^{-1}B$
   Since multiplying a matrix by its inverse yields the Identity matrix ($A^{-1}A = I$), the left side becomes $IX$, which is simply $X$: $X = A^{-1}B$

Solved Examples

• A Unique Solution. Suppose we have the following system of equations:

$$\begin{cases} x + y + z = 6 \\ 2y + 5z = -4 \\ 2x + 5y - z = 27 \end{cases}$$

In matrix form, this is, $\begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 5 \\ 2 & 5 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \\ 27 \end{pmatrix}$.

To solve for $X$, follow these steps:

1. Calculate the determinant of A. $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = 1(2(-1) - 5(5)) - 1(0(-1) - 5(2)) + 1(0(5) - 2(2)) = 1(-27) - 1(-10) + 1(-4) = -27 + 10 - 4 = -21$.
   Because the determinant is not zero, a unique solution exists.
2. Find the Cofactor matrix C. Using the formula $C_{ij} = (-1)^{i+j} \times \text{minor}(A_{ij})$, we get: $C = \begin{pmatrix} -27 & 10 & -4 \\ 6 & -3 & -3 \\ 3 & -5 & 2 \end{pmatrix}$
3. Transpose the Cofactor matrix to obtain the Adjugate matrix $\text{adj}(A) = C^T = \begin{pmatrix} -27 & 6 & 3 \\ 10 & -3 & -5 \\ -4 & -3 & 2 \end{pmatrix}$
4. Calculate the inverse $A^{-1}$: $A^{-1} = \frac{1}{\det(A)} \times \text{adj}(A)$.
5. Solve for $X$ by multiplying $A^{-1}B$:: $X = \frac{-1}{21} \begin{pmatrix} -27 & 6 & 3 \\ 10 & -3 & -5 \\ -4 & -3 & 2 \end{pmatrix} \begin{pmatrix} 6 \\ -4 \\ 27 \end{pmatrix}$
   $X = \frac{-1}{21} \begin{pmatrix} (-27)(6) + (6)(-4) + (3)(27) \\ (10)(6) + (-3)(-4) + (-5)(27) \\ (-4)(6) + (-3)(-4) + (2)(27) \end{pmatrix} = \frac{-1}{21} \begin{pmatrix} -162 - 24 + 81 \\ 60 + 12 - 135 \\ -24 + 12 + 54 \end{pmatrix} = \frac{-1}{21} \begin{pmatrix} -105 \\ -63 \\ 42 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \\ -2 \end{pmatrix}$
6. Conclusion. Therefore, The solution to our original system of equation is $x = 5$, $y = 3$, $z = -2$.
   Geometrically, this means our three equations represent three flat planes in 3D space that all intersect perfectly at the single point $(5, 3, -2)$.
   🧩 This method provides a systematic way to solve systems of linear equations using matrix operations, making it a powerful tool in linear algebra.

• A Singular Matrix (Infinitely Many Solutions). Suppose we have the following system of equations:

$\begin{cases} x + y + z = 7 \\ 3x - 2y - z = 4 \\ x + 6y + 5z = 24 \end{cases}$

In matrix form $AX = B$, this is, $\begin{pmatrix} 1 & 1 & 1 \\ 3 & -2 & -1 \\ 1 & 6 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \\ 24 \end{pmatrix}$.

1. Calculate the determinant of A. $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) = 1((-2)(5) - (-1)(6)) - 1(3(5) - (-1)(1)) + 1(3(6) - (-2)(1)) = 1(-10 + 6) - 1(15 + 1) + 1(18 + 2) = 1(-4) - 1(16) + 1(20) = -4 - 16 + 20 = 0$.
   Because $\det(A) = 0$, the matrix $A$ is singular. The inverse $A^{-1}$ does not exist. This means the system either has no solutions (inconsistent) or infinitely many solutions (dependent).
   To find out which it is, we must abandon the inverse method and use Gaussian Elimination (Row Reduction) on the augmented matrix.
2. Convert to Row Echelon Form. A matrix is said to be in row echelon form when all its non-zero rows have a pivot, that is, a non-zero entry such that all the entries to its left and below it are equal to zero.
3. Start with the augmented matrix: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\\ 3 & -2 & -1 &\bigm| & 4\\\ 1 & 6 & 5 &\bigm| & 24\end{smallmatrix}\Biggr)$
   Eliminate the x terms from the second and third rows, $R_2 \leftarrow R_2 - 3R_1$ and $R_3 \leftarrow R_3 - R_1$: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\\ 0 & -5 & -4 &\bigm| & -17\\\ 0 & 5 & 4 &\bigm| & 17\end{smallmatrix}\Biggr)$
   Eliminate the y term from the third row, $R_3 \leftarrow R_3 + R_2$: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\\ 0 & -5 & -4 &\bigm| & -17\\\ 0 & 0 & 0 &\bigm| & 0\end{smallmatrix}\Biggr)$.
4. The final augmented matrix shows that we have a row of all zeros, indicating that the system is consistent (0x+0y+0z = 0 ↭ 0 = 0 is a true statement), but we have a “free variable”. Geometrically, these three planes don't meet at a single point —they intersect along an infinite line.
5. Parametrize the Solution. Translate the remaining rows back into equations: (1) $x + y + z = 7$. (2) $-5y - 4z = -17$.
   Let’s solve for $y$ and $z$ in terms of $x$. First, express $y$ and $z$ relationships: From equation (2), we get $y = \frac{-17 + 4z}{-5}$
   Substitute this $y$ into equation (1): $x + \frac{-17 + 4z}{-5} + z = 7$. Multiply by 5 to clear the fraction: $5x + 17 - 4z + 5z = 35$. Simplify, $5x + z = 18 \implies z = -5x + 18$
   Now substitute $z$ back into our $y$ equation: $y = \frac{-17 + 4(-5x + 18)}{-5} = \frac{-17 - 20x + 72}{-5} = \frac{-20x + 55}{-5} = 4x - 11$
6. Conclusion: The system has infinitely many solutions. For any real number $x$, the solution coordinate is $(x, 4x - 11, -5x + 18)$.

• Suppose we have the following system of equations:

$\begin{cases} x - 3y + z = 4 \\ -x + 2y - 5z = -4 \\ 5x - 13y + 13z = 8 \end{cases}$

In matrix form, this can be written as $AX = B$: $\begin{pmatrix} 1 & -3 & 1 \\ -1 & 2 & -5 \\ 5 & -13 & 13 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \\ 8 \end{pmatrix}$

To solve for $X$, we ideally need to calculate the inverse matrix, $A^{-1}$. However, we must first check if the inverse exists.

1. Calculate the determinant of A. Using the standard formula for a $3 \times 3$ matrix determinant: $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$.
   $\det(A) = 1(2 \cdot 13 - (-5)(-13)) - (-3)(-1 \cdot 13 - (-5)(5)) + 1(-1(-13) - 2(5)) = 1(26 - 65) + 3(-13 + 25) + 1(13 - 10) = 1(-39) + 3(12) + 1(3) = -39 + 36 + 3 = 0$.
   Because the determinant is $0$, the matrix $A$ is singular. This tells us immediately that the system cannot have a single unique solution —it must either be inconsistent (no solutions) or have infinitely many solutions.
2. Convert to Row Echelon Form. To find out which is the case, we use Gaussian elimination. A matrix is said to be in row echelon form when all its non-zero rows have a pivot, that is, a non-zero entry such that all the entries to its left and below it are equal to zero.
   Start with the augmented matrix: $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\\ -1 & 2 & -5 &\bigm| & -4\\\ 5 & -13 & 13 &\bigm| & 8 \end{smallmatrix}\Biggr)$
   Eliminate the x term from row 2: $R_2 \to R_2 + R_1$ (Add row 1 to row 2): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\\ 0 & -1 & -4 &\bigm| & 0\\\ 5 & -13 & 13 &\bigm| & 8 \end{smallmatrix}\Biggr)$
   Eliminate the $x$ term from row 3: $R_3 \to R_3 - 5R_1$ (Subtract 5 times row 1 from row 3): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\\ 0 & -1 & -4 &\bigm| & 0\\\ 0 & 2 & 8 &\bigm| & -12 \end{smallmatrix}\Biggr)$
   Eliminate the y term from the third row: $R_3 \to R_3 + 2R_2$ (Multiply row 2 by 2 and add to row 3): $\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\\ 0 & -1 & -4 &\bigm| & 0\\\ 0 & 0 & 0 &\bigm| & -12 \end{smallmatrix}\Biggr)$
3. Identify Inconsistency. The third rows indicates or algebraically translates to $0x +0y + 0z = -12$, which simplifies to $0 = -12$. This is a mathematical contradiction, which means the system of equations is inconsistent. There are no solutions to this system.

Equations of planes

A plane in three-dimensional space can be thought of as a flat, two-dimensional surface that extends infinitely in all directions.

To define a plane mathematically, we need two things:

1. A point on the plane, often denoted as $P_0 = (x_0, y_0, z_0)$.
2. A normal vector to the plane, denoted as $\vec{N} = \langle a, b, c \rangle$. This vector sticks straight out of the plane and is completely perpendicular (orthogonal) to every vector that lies on the plane’s surface.

Deriving the Equation:

Given a point $P_0 = (x_0, y_0, z_0)$ and a normal vector $\vec{N}$, we can find the equation of the plane. Here’s how:

1. Vectors in the plane. Any arbitrary point $P = (x, y, z)$ on the plane creates a vector $\vec{P_0P}$ with the point $P_0$. This vector is given by $\vec{P_0P} = \langle x - x_0, y - y_0, z - z_0 \rangle$.
2. Perpendicularity Condition. The vector $\vec{N}$ is perpendicular to the plane, it’s also perpendicular to any vector that lies in the plane. Mathematically, this means that their dot product is zero: $\vec{P_0P} \cdot \vec{N}=0$.
3. Dot Product Calculation. Expanding this yields the scalar equation of the plane: $\langle x - x_0, y - y_0, z - z_0 \rangle \cdot \langle a, b, c \rangle = 0$
   $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

 

General Form of the Plane Equation

This scalar equation can be distributed and rearranged into the standard general form: $ax + by + cz = d$ where $d = ax_0 + by_0 + cz_0$ and $\vec{N} = \langle a, b, c \rangle$.

Key insight: The coefficients of $x, y,$ and $z$ in the general form of a plane’s equation always correspond directly to the components of its normal vector $\vec{N}$.

Solved exercises

• Plane Through the Origin. Find the equation of a plane through the origin with normal vector $\vec{N} = \langle 1, 5, 10 \rangle$ (Figure viii)

1. Understanding the Normal Vector. A normal vector to a plane is a vector that is perpendicular to the surface of the plane. $\vec{N} = \langle 1, 5, 10 \rangle$.
2. Point on the plane. Since the plane passes through the origin, the point we use is $P_0 = (0, 0, 0)$.
3. Equation of the plane. We use the fact that any point $P(x,y,z)$ on the plane must satisfy the condition that the vector from the origin $P_0 = (0, 0, 0)$ to P is perpendicular to the normal vector the plane: $\vec{OP} \cdot \vec{N} = 0 \iff \langle x, y, z \rangle \cdot \langle 1, 5, 10 \rangle = 0$.
4. Calculating the dot product: $\mathbf{x + 5y + 10z = 0}$.

 

• Line Perpendicular to a Plane. Find the vector equation of the line passing through point $P(2, -4, 3)$ that is perpendicular to the plane $x + 4y - 2z = 5$.

1. Identify the normal vector of the plane. The normal vector $\vec{N}$ of the plane $x + 4y - 2z = 5$ can be directly obtained from the coefficients of x, y, and z in the plane equation. Thus,$\vec{N} = \langle 1, 4, -2 \rangle$. Recall that the normal vector is perpendicular to the plane.

2. Set up the line equation. The line passing through the point P(2, -4, 3) and in the direction of $\vec{N}$ can be written using the point-direction form of the line equation. If $\vec{r_0}$ is the position vector of the point P and $\vec{d}$ is the direction vector (which, in our particular case, is the normal vector $\vec{N}$), the vector equation of the line is (illustration ii) $\vec{r(t)}=\vec{r_0}+t\vec{d} = \vec{r_0}+t\vec{N}$.

3. Substitute the known values: $\vec{r}(t) = \langle 2, -4, 3 \rangle + t\langle 1, 4, -2 \rangle$

4. Conclusion: The vector equation of the line passing through point $P(2, -4, 3)$ that is perpendicular to the plane $x + 4y - 2z = 5$ is $\mathbf{\vec{r}(t) = \langle 2+t, -4+4t, 3-2t \rangle}$.

   

• Plane Through a Specific Point. Find the equation of a plane through $P_0 = (2, 1, -1)$ with normal vector $\vec{N} = \langle 1, 5, 10 \rangle$.

1. Understanding the Normal Vector. A normal vector to a plane is a vector that is perpendicular to the surface of the plane. $\vec{N} = \langle 1, 5, 10 \rangle$
2. Point on the plane: $P_0 = ⟨2, 1, -1⟩$.
3. Equation of the plane. We use the fact that any point P on the plane must satisfy the condition that the vector from $P_0$ to P, $\vec{P_0P}$, is perpendicular to the normal vector the plane: $\vec{P_0P} \cdot \vec{N} = 0 \iff \langle x -x_0, y -y_0, z-z_0 \rangle \cdot \langle a, b, c \rangle$.
   Calculating the dot product: $a(x-x_0)+b(y-y_0)+c(z-z_0) = 0$
4. Substitute the given values: $1(x-2) +5(y-1) +10(z+1)=0$.
5. Expand and simplify: $x −2 +5y −5 +10z +10 = 0 \iff [\text{Combine like terms}] x +5y +10z +3 = 0 \iff [\text{Rearrange the equation to standard form:}] x + 5y + 10z = -3.$

Pro-Tip: Notice that if we are given the equation of a plane in this form, we can quickly get our normal vector, namely $\vec{N} = \langle 1, 5, 10 \rangle$ or alternatively, if know our normal vector, we can write down the left-hand side of the plane’s equation, and we plug in $P_0 = ⟨2, 1, -1⟩$ in the left-hand side to get the value of the right-hand side of the plane’s equation, $1(2) + 5(1) + 10(-1) = -3$.

• Parallel or Perpendicular?. Are the vector $\vec{v} = \langle 1, 2, -1 \rangle$ and the plane $x + y + 3z = 5$ parallel, perpendicular, or neither?

1. Extract the Normal Vector from the Plane Equation. The equation of the plane is given as $x + y + 3z = 5$. This is the scalar equation of the plane. From the plane equation $x + y + 3z = 5$, we can directly extract the normal vector $\vec{N}$ that is perpendicular to the plane. Its components are given by the coefficients of x, y, and z in the plane equation: $\vec{N} = \langle 1, 1, 3 \rangle$.
2. Calculate the dot product of $\vec{v}$ and $\vec{N}$: $\vec{v} \cdot \vec{N}$ =[Recall, $\vec{a} \cdot \vec{b}=a_1b_1+a_2b_2+a_3b_3$] $ (1)(1) + (2)(1) + (-1)(3) = 1 + 2 - 3 = 0.$
3. Interpret the result. To determine if $\vec{v}$ is parallel, perpendicular, or neither to the plane, we will interpret the dot product result of $\vec{v}$ and $\vec{N}$. If the dot product is zero, it indicates that $\vec{v}$ and $\vec{N}$ are perpendicular to each other. Since $\vec{v}$ is perpendicular to $\vec{N}$, and $\vec{N}$ is the normal vector to the plane, then $\vec{v}$ must lie completely flat against (or alongside) the plane. Therefore, $\vec{v}$ is parallel to the plane..

• Plane Given Three Points. Find an equation for the plane passing through $Q(-1, 1, 2)$, $R(-4, 2, 2)$, and $S(-2, 1, 5)$ (Figure iii).

1. Determine two vectors in the plane. Since the plane passes through points Q, R, and S, then we can easily find two vectors $\vec{QR} = R - Q = \langle -4 - (-1), 2 - 1, 2 - 2 \rangle = \langle -3, 1, 0 \rangle$ and $\vec{QS} = S - Q = \langle -2 - (-1), 1 - 1, 5 - 2 \rangle = \langle -1, 0, 3 \rangle$ that lie in the plane.

2. Calculate the Normal Vector to the plane. We need a vector perpendicular to both $\vec{QR}$ and $\vec{QS}$. We find this by taking their cross product $\vec{QR} \times \vec{QS}$:
   $\vec{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 1 & 0 \\ -1 & 0 & 3 \end{vmatrix} = \mathbf{i}(3 - 0) - \mathbf{j}(-9 - 0) + \mathbf{k}(0 - (-1)) = 3\mathbf{i} + 9\mathbf{j} + \mathbf{k} = \langle 3, 9, 1 \rangle$

3. Determine the plane equation. Let P(x, y, z) be an arbitrary point P in the plane, the vector $\vec{QP} = \langle x +1, y -1, z -2 \rangle$ lie in the plane, and therefore is perpendicular to $\vec{N} = \langle 3, 9, 1 \rangle$. Therefore, their dot product must be zero: $\vec{N}\cdot \vec{QP} = 0 \iff \langle 3, 9, 1 \rangle \cdot \langle x +1, y -1, z -2 \rangle = 0$.
   Calculate the dot product: $3(x+1)+9(y-1)+(z-2) = 0$
   Simplify the equation: $3x +3 +9y −9 +z −2 = 0 \iff 3x + 9y + z = 8$.

4. Conclusion. $3x + 9y + z = 8$ is the plane that passes through Q, R, and S.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
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