It is forbidden to kill; therefore all murderers are punished unless they kill in large numbers and to the sound of trumpets, Voltaire
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., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.
The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a_{1}+b_{1}), (a_{2}+b_{2}), (a_{3}+b_{3}) >.
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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a_{1}-b_{1}), (a_{2}-b_{2}), (a_{3}-b_{3}) >.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.
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. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$
It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$
The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input 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 following formula, $\vec{A}x\vec{B} = det\Bigl(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}\Bigr) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
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 · X = U provides a more convenient and concise notation and also an efficient way to solve system of linear equations. A is a 3 x 3 matrix representing the coefficients, X is a 3 x 1 column vector representing the variables and U is a 3 x 1 column vector representing the results.
Given a square matrix A (it is 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)≠ 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.
In linear algebra, solving a system of linear equations can be efficiently done using matrices.
Consider a system of linear equations:
$\begin{cases} a_{11}x_1 +a_{12}x_2 + ···a_{1n}x_n = b_1 \\ a_{21}x_1 +a_{22}x_2 + ···a_{2n}x_n = b_2 \\ . \\ . \\ . \\ a_{n1}x_1 +a_{n2}x_2 + ···a_{nn}x_n = b_n \end{cases}$
It can be written or represented in matrix form as: AX = B, where:
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.
To solve AX = B, follow these steps:
$\begin{cases} x + y + z = 6 \\ 2y + 5z = -4 \\ 2x + 5y -z = 27 \end{cases}$
In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & 1 & 1\\ 0 & 2 & 5\\ 2 & 5 & -1\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}6\\ -4\\ 27\end{smallmatrix}\Bigr)$.
To solve for X, we need to calculate A^{-1}.
Therefore, The solution to our original system of equation is X = ⟨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.
$\begin{cases} x + y + z = 7 \\ 3x - 2y -z = 4 \\ x + 6y +5z = 24 \end{cases}$
In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & 1 & 1\\ 3 & -2 & -1\\ 1 & 6 & 5\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}7\\ 4\\ 24\end{smallmatrix}\Bigr)$.
To solve for X, we need to calculate A^{-1}.
If A is singular (det(A) = 0), we cannot find A^{-1}. Instead, we need to use other methods like row reduction to understand the nature of the solutions.
Start with the augmented matrix: $\Biggl(\begin{smallmatrix}1 & 1 & 1 &\bigm| & 7\\ 3 & -2 & -1 &\bigm| & 4\\ 1 & 6 & 5 &\bigm| & 24\end{smallmatrix}\Biggr)$
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) and has free variables.
$\begin{cases} x + y + z = 7 (i) \\ -5y -4z = -17 (ii) \end{cases}$
-5y -4z = -17 (ii)⇒$ y = \frac{-17+4z}{-5}$⇒[Substituting y into the first equation (i)] $x + \frac{-17+4z}{-5} +z = 7$ ⇒[Multiply through by 5 to clear the fraction] $5x +17 -4z + 5z = 35 ⇒ 5x + z = 35-17 ⇒ z = 18-5x$
$ y = \frac{-17+4z}{-5} = \frac{-17+4(18-5x)}{-5} = \frac{-17+72-20x}{-5} = \frac{55-20x}{-5} = -11+4x$
The system has infinitely many solutions of the form ⟨x, 4x -11, -5x +18⟩.
$\begin{cases} x -3y + z = 4 \\ -x + 2y -5z = -4 \\ 5x - 13y +13z = 8 \end{cases}$
In matrix form, this is, $\Bigl(\begin{smallmatrix}1 & -3 & 1\\ -1 & 2 & -5\\ 5 & -13 & 13\end{smallmatrix}\Bigr)\Bigl(\begin{smallmatrix}x\\ y\\ x\end{smallmatrix}\Bigr) = \Bigl(\begin{smallmatrix}4\\ -4\\ 8\end{smallmatrix}\Bigr)$.
To solve for X, we need to calculate A^{-1}.
Start with the augmented matrix:
$\Biggl(\begin{smallmatrix}1 & -3 & 1 &\bigm| & 4\\ -1 & 2 & -5 &\bigm| & -4\\ 5 & -13 & 13 &\bigm| & 8 \end{smallmatrix}\Biggr)$
Identify Inconsistency. The third rows indicates 0x +0y + 0z = -12 ⇒ 0 = -12. This is a contradiction, which means the system of equations is inconsistent. Therefore, there are no solutions to this system.
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:
Given a point P_{0} and a normal vector $\vec{N}$, we can find the equation of the plane. Here’s how:
The scalar equation of the plane can be rearranged into a more familiar form: ax + by + cz = d where d = ax_{0} + by_{0} + cz_{0}, ⟨a, b, c⟩ = $\vec{N}$.
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, $\vec{N} = ⟨1, 4, -2⟩$. Recall that the normal vector is perpendicular to the plane.
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) $\vec{r(t)}=\vec{r_0}+t\vec{d} = \vec{r_0}+t\vec{N}$.
Substitute the known values: $\vec{r(t)} = ⟨2, -4, 3⟩ + t⟨1, 4, -2⟩ = ⟨2+t, -4+4t, 3-2t⟩.$
In conclusion, the vector equation of the line passing through the point P(2, -4, 3) and perpendicular to the plane x +4y -2z = 5 is $\vec{r(t)}=⟨2+t, -4+4t, 3-2t⟩.$
Notice that if we are given the equation of a plane in this form, we can quickly get our normal vector, namely $\vec{N} = ⟨1, 5, 10⟩$ 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} in the left-hand side to get the value of the right-hand side of the plane’s equation.
Determine two vectors in the plane. Since the plane passes through points Q, R, and S, then we can find two vectors $\vec{QR}= R−Q = ⟨−4−(−1),2−1,2−2⟩ = ⟨-3, 1, 0⟩$ and $\vec{QS} = S−Q = ⟨−2−(−1),1−1,5−2⟩ = ⟨-1, 0, 3⟩$ that lie in the plane.
Calculate the Normal Vector to the plane. The normal vector $\vec{N}$ to the plane can be found by taking the cross product of $\vec{QR}$ and $\vec{QS}$
$\vec{N} = |\begin{smallmatrix} \vec{i} & \vec{j} & \vec{k}\\ -3 & 1 & 0 \\ -1 & 0 & 3 \end{smallmatrix}| = 3\vec{i} +9\vec{j} + \vec{k} = ⟨3, 9, 1⟩$
Determine the plane equation. Let P(x, y, z) be an arbitrary point P in the plane, the vector $\vec{QP} = ⟨x +1, y -1, z -2⟩$ lie in the plane, and therefore is perpendicular to $\vec{N}$. Therefore, their dot product must be zero: $\vec{N}·\vec{QP} = 0 ↭ ⟨3, 9, 1⟩·⟨x +1, y -1, z -2⟩ = 0$ ↭[Calculate the dot product:] 3(x+1)+9(y-1)+(z-2) = 0 ↭[Simplify the equation:] 3x +3 +9y −9 +z −2 = 0 ↭ 3x + 9y + z = 8. Hence, 3x + 9y + z = 8 is the plane that passes through Q, R, and S.