Logic will get you from A to B. Imagination will take you everywhere, Albert Einstein.
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, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}\bigl) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
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,
$\begin{cases} 2x_1 +3x_2 + 3x_3 = u_1 \\ 2x_1 +4x_2 + 4x_3 = u_2 \\ x_1 + x_2 + 2x_3 = u_3 \end{cases}$
$(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}\bigl) (\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}\bigl) = (\begin{smallmatrix}u_1\\ u_2\\u_3\end{smallmatrix}\bigl)$, A · X = U provides a more convenient and concise notation and an efficient way to solve system of linear equations where A is a 3 x 3 matrix representing coefficients, X is a 3 x 1 column vector representing the results where we basically do dot products between the rows of A (a 3 x 3 matrix) and the column vector of X (a 3 x 1 matrix).
In geometry, understanding the relationship between points , lines, and planes is fundamental. A plane can be thought of as the intersection of three-dimensional figures.
A point is the intersection of two lines; a line is the intersection of two planes; and a plane can be thought as the intersection of three-dimensional figures.
Unlike in ℝ^{3} where the equation y = mx + b describe a line, in ℝ^{3}, this equation describes a plane.
A line in three-dimensional space can be visualized as the path or trajectory of a moving point. Imagine a point Q(t) that moves along the line. At t = 0, it is at Q_{0}, at t = 1 it is at Q_{1}, and at any time t, it is at Q_{t}. Besides, $\vec{Q_0Q(t)}=t\vec{Q_0Q_1}$ because it moves at a constant speed.
Let’s break it down with an example:
Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}2\\ -1\\ 3\end{smallmatrix}\bigr)$
This results in the parametric equations:
$\begin{cases} x(t) = - 1 + 2t \\ y(t) = 2 + t \\ z(t) = 2 -3t \end{cases}$
First, Q_{0} $(x+2y+4z)\bigg|_{(-1, 2, 2)} = −1 + 2⋅2 + 4⋅2 = -1 + 4 + 8 = 11$ > 7 and Q_{1} = 1 + 2·3 + 4·(-1) =3 < 7 are not in the plane, so Q_{0} and Q_{1} are on opposite sides of the plane. Therefore, we can conclude that the line intersect the plane between these points.
To know the exact point where the line intersect the plane, we just need to plug the line’s equation (substitute the parametric equations of the line) in the plane’s equation, x(t) + 2y(t) + 4z(t) = 7 ↭ (-1 + 2t) + 2(2 + t) + 4(2 -3t) = 7 ↭[Simplifying] -8t + 11 = 7 ↭ -8t = -4 ↭ 2t = 1. Q(t) is in the plane ↭ t = ^{1}⁄_{2}, i.e., halfway Q_{0} and Q_{1}, hence the line intersects the plane x + 2y + 4z = 7 at t = $Q(\frac{1}{2}) = (0, \frac{5}{2},\frac{1}{2})$.
Verification: $(x+2y+4z)\bigg|_{(0, \frac{5}{2},\frac{1}{2})} = 5 + 2 = 7.$
Line Intersects the Plane. If the parametric equations of the line, when substituted into the plane equation, yield a specific value of the parameter t, then the line intersects the plane at that value of t , e.g., Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}2\\ -1\\ 3\end{smallmatrix}\bigr)$ intersects the plane x + 2y + 4z = 7 at t = $Q(\frac{1}{2}) = (0, \frac{5}{2},\frac{1}{2})$.
Line lies in the plane: If, after substituting the parametric equations into the plane equation, the result simplifies to a true statement (like 7=7) for all values of t, the line lies entirely within the plane.
Suppose we choose a point P_{0} on the plane x + 2y + 4z = 7, say P_{0} = (1, 1, 1), and a direction vector $\vec{d}$ that is orthogonal to the normal vector of the plane (hence, it is parallel to the plane) $\hat{\mathbf{n}} = ⟨1, 2, 4⟩$, e.g., $\vec{d} = ⟨0, -2, 1⟩$: ⟨0, -2, 1⟩·⟨1, 2, 4⟩ = 0 -4 + 4 = 0.
The line Q(t) = $\bigl(\begin{smallmatrix}1\\ 1\\ 1\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}0\\ -2\\ 1\end{smallmatrix}\bigr)$ lies in the plane x + 2y + 4z = 7. Verification: 1 +2(1 -2t) + 4(1 +t) = 7 ↭ 1 +2 -4t +4 + 4t = 7 ↭ 7 = 7. Since the equation holds true for all t, the line lies entirely in the plane.
Line is Parallel to the Plane. If, after substitution, the result is a contradiction (like a non-zero constant equal to zero), the line is parallel to the plane and does not intersect it.
Suppose we choose a point P_{0} that is not the plane x + 2y + 4z = 7, say P_{0} = (-1, 2, 2), and a direction vector $\vec{d}$ that is orthogonal to the normal vector of the plane (hence, it is parallel to the plane) $\hat{\mathbf{n}} = ⟨1, 2, 4⟩$, e.g., $\vec{d} = ⟨0, -2, 1⟩$: ⟨0, -2, 1⟩·⟨1, 2, 4⟩ = 0 -4 + 4 = 0.
The line Q(t) = $\bigl(\begin{smallmatrix}-1\\ 2\\ 2\end{smallmatrix}\bigr)+t\bigl(\begin{smallmatrix}0\\ -2\\ 1\end{smallmatrix}\bigr)$ is parallel to the plane x + 2y + 4z = 7. Verification: -1 + 2(2 -2t) + 4(2 + t) = 7 ↭ - 1 + 4 -4t + 8 + 4t = 7 ↭ 11 = 7 ⊥ Since the equation results in a contradiction, this confirms that the line is parallel to the plane and does not intersect it.
A parametric equation defines a group of quantities as functions of one or more independent variables called parameters. In simpler terms, it is a set of equations where each equation describes one of the coordinates (or other dependent variables) in terms of an independent variable, known as the parameter, e.g., x = x(t), y = y(t), that defines or specifies the (dependent) variables x and y as functions of a parameter or independent variable t.
Parametric equations are often used to describe motions and paths in the plane or in space, where the x, y, and z coordinates vary with some other parameter, such as t (time) or even θ (for example, a particular angle).
Sketch the curves described by the following parametric equations:
x = t + 2, y = t^{2}, -2 ≤ t ≤ 2 (Illustration B). For t = −2, x =−2+2=0 and y= (-2)^{2}=4. For t = -1, x = 1, y = 1. For t = 0, x = 2, y = 0. For t = 1, x = 2, y = 1. For t = 2, x = 4, y = 4. Plot these points and connect them smoothly to see the parabolic shape. The parabola opens upwards and shifts rightward by 2 units along the x-axis.
x = $\sqrt{t}$, y = 3t + 1, t ≥ 0 (Illustration C). For t = 0, x = 0, y = 1. For t = 1, x = 1, y = 4. For t = 4, x = 2, y = 13. For t = 9, x = 3, y = 28. The curve starts at (0,1) and moves upwards steeply as t increases, forming a rightward-opening shape.
x = 4cos(t), y = 4sin(t), 0 ≤ t ≤ 2π, (Illustration D). This is the graph of a circle with radius 4 centered at the origin, moving in a counterclockwise orientation.
Eliminating the parameter involves rewriting parametric equations as a single equation in terms of x and y. This process can help you understand the graph of a curve more clearly by relating the variables directly without the intermediary parameter.
Let’s work through the examples provided and clarify each step.
Solve one of the equation for t. From (i), $t = \frac{x}{2}$, then substitute into the other equation (ii), $y = t^2 = (\frac{x}{2})^2 = \frac{x^2}{4}$, and this is the Cartesian form, $y = \frac{x^2}{4}$. This Cartesian form represents a parabola opening upwards.