Mathematical Foundations and Vector Applications: Lines, Planes, and Force Systems

Coordinate Systems and Linear Equations in 2D

• Transitioning Dimensions: The discussion begins with the standard two-dimensional $x$ and $y$ coordinates (Cartesian system), moves into three-dimensional space ($3D$), and will eventually extend to $n$ dimensions ($R^n$).

• General Form of a Linear Equation: The standard algebraic form of a line in an $x$ $y$ plane is given by the general form:     $ax + by + c = 0$

  • Multiple names exist for the same linear relationship depending on its representation: vector form, general form, and normal form.

• Slope-Intercept Form: If $b \neq 0$, the general form can be solved for $y$ to produce the slope-intercept form:     $by = -ax + c$     $y = -\frac{a}{b}x + \frac{c}{b}$

  • The term $-\frac{a}{b}$ represents the slope ($m$).

  • The term $k = \frac{c}{b}$ represents the $y$-intercept, specifically the point $(0, k)$.

• Vertical Lines and Undefined Slopes: If $b = 0$, the slope becomes undefined (approaching infinity). For example, the equation $x = 2$ represents a vertical line passing through $2$ on the $x$-axis; such cases are technically outside the standard slope-intercept conversion.

Vector and Normal Forms of Lines

• The Normal Vector (n): In the equation of a line, the coefficients of $x$ and $y$ can be represented as a vector $n = \begin{pmatrix} a \\ b \end{pmatrix}$. This vector is called the normal, and it is perpendicular to the line.

• Dot Product Representation: The left-hand side of a linear equation can be conceptualized as a dot product between the normal vector $n$ and a generic vector $x = \begin{pmatrix} x \\ y \end{pmatrix}$.

• Lines Through the Origin: For a line passing through the origin ($0, 0$), the equation is:     $n \cdot x = 0$

• Orthogonality Proof: If a line is $y = -2x$ (or $2x + y = 0$), the slope is $-2$. The perpendicular slope (negative reciprocal) is $\frac{1}{2}$. The normal vector is $n = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$. Any vector on the line (like $x = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$) will have a dot product of zero with the normal:     $(2 \times 1) + (1 \times -2) = 0$

• Particle Metaphor: A line can be viewed as the path of a particle moving over time ($t$). For example, if a particle at the origin moves at one unit per second along the line, at $t = 1$, it is at point $(1, -2)$; at $t = 2$, it is at $(2, -4)$. Measures along the line are referred to as parameters.

Lines Involving Shifts (y-intercepts)

• Equation with Non-Zero Constant: Consider the line $2x + y = 5$. To write the normal form:     $n \cdot (x - p) = 0$     $n \cdot x = n \cdot p$

  • $n$ is the normal vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

  • $p$ is a specific point on the line (e.g., the y-intercept $(0, 5)$).

  • $n \cdot p = (2 \times 0) + (1 \times 5) = 5$.

• Vertical Shifts: Changing the constant from $0$ to $5$ represents a vertical shift of exactly five units up. The slope and the normal vector remain identical to the line passing through the origin; only the position in space changes.

Vector and Parametric Equations of Lines

• Vector Form: This form is applicable in both $R^2$ and $R^3$ (and higher):     $x = p + td$

  • $p$: A position vector for a specific point on the line.

  • $t$: A scalar parameter.

  • $d$: The direction vector (parallel to the line).

• Parametric Form: Derived by breaking the vector form into coordinate components.

• Example: Line in $R^3$:

  • Point $p = (1, 2, -1)$.

  • Direction vector $d = (5, -1, 3)$.

  • Vector Form: $x = (1, 2, -1) + t(5, -1, 3)$.

  • Parametric Equations:         $\begin{cases} x = 1 + 5t \\ y = 2 - t \\ z = -1 + 3t \end{cases}$

  • Verification: At $t = 0$, the coordinates return the initial point $(1, 2, -1)$.

• Line Determined by Two Points: To find the equation of a line passing through points $P$ and $Q$:

  1. Use point $P$ as the position vector $p$.

  2. Calculate the direction vector $d$ by subtracting the coordinates: $d = Q - P$.

  3. Example with $P(-5, -1, 5, 0)$ (implied as 4D or 3D context) and $Q(2, 1, 1)$. For points $Q(2, 1, 1)$ and $P(-1, 5, 0)$, $d = (2 - (-1), 1 - 5, 1 - 0) = (3, -4, 1)$.

Planes in 3D Space

• General Form of a Plane: In three dimensions, the equation $ax + by + cz = d$ represents a plane, not a line.

• Normal Vector of a Plane: The vector $n = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ is perpendicular to every vector lying within that plane.

• Normal Form of a Plane: $n \cdot (x - p) = 0$.

• Example: Finding Equation of a Plane:

  • Point: $(6, 0, 1)$.

  • Normal vector $n = (1, 2, 3)$.

  • Calculation: $1(x) + 2(y) + 3(z) = (1 \times 6) + (2 \times 0) + (3 \times 1)$.

  • Result: $x + 2y + 3z = 9$.

• Vector Form of a Plane: A plane requires two non-parallel direction vectors ($u$ and $v$) and a point $p$:     $x = p + su + tv$

• Parametric Form of a Plane: Derived from the vector form, requiring two parameters ($s$ and $t$) because a plane is a two-dimensional subspace of $3D$ space.

  • Example: Point $p(6, 0, 1)$. Points on plane found by trial and error: $Q(9, 0, 0)$ and $R(3, 3, 0)$.

  • Vectors: $u = Q - p = (3, 0, -1)$; $v = R - p = (-3, 3, -1)$.

  • Parametric Equations:         $\begin{cases} x = 6 + 3s - 3t \\ y = 3t \\ z = 1 - s - t \end{cases}$

Metric Calculations: Distance and Projection

• Distance from a Point to a Line: To find the distance from point $B$ to a line passing through point $A$ with direction $d$:

  1. Create vector $v = AB = B - A$.

  2. Calculate the projection of $v$ onto $d$: $\text{proj}_d(v) = \frac{v \cdot d}{d \cdot d}d$.

  3. The distance is the magnitude of the vector resulting from $v - \text{proj}_d(v)$.

• Distance from a Point to a Plane: The distance of a point $(x_0, y_0, z_0)$ to a plane $ax + by + cz = d$ can be calculated via the formula:     $\text{distance} = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$

  • This is conceptually the magnitude of the projection of vector $v$ (from a point on the plane to the external point) onto the normal vector $n$.

Physics Applications of Vectors (Section 1.4)

• Resultant Forces: Combining multiple forces into a single vector.

  • Example 1: Force $F_1 = 12\,N$ (North) and $F_2 = 5\,N$ (East).

  • Resultant Magnitude ($R$): Using the Pythagorean theorem: $\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,N$.

• Resolving Forces: Breaking a single force into perpendicular components.

  • Example 2: Force of $10\,N$ at a $60^\circ$ angle.

  • Horizontal component: $10 \times \cos(60^\circ) = 10 \times 0.5 = 5\,N$.

  • Vertical component: $10 \times \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\,N$.

• Block on an Inclined Plane:

  • Block mass: $10\,kg$. Gravity ($g$): $9.8\,m/s^2$. Weight ($W$): $10 \times 9.8 = 98\,N$.

  • Ramp angle: $30^\circ$.

  • Force parallel to ramp (sliding force): $W \times \sin(30^\circ) = 98 \times 0.5 = 49\,N$.

• Tension in Wires: A painting ($20\,kg$) suspended by two wires at angles of $30^\circ$ and $45^\circ$. Weight $W = 20 \times 9.8 = 196\,N$.

  • Method 1 (System of Equations):         Horizontal equilibrium: $T_1\cos(30^\circ) = T_2\cos(45^\circ)$.         Vertical equilibrium: $T_1\sin(30^\circ) + T_2\sin(45^\circ) = 196$.

  • Method 2 (Law of Sines):         $\frac{196}{\sin(75^\circ)} = \frac{T_1}{\sin(45^\circ)} = \frac{T_2}{\sin(60^\circ)}$         Calculated Tensions: $T_1 \approx 143.4\,N$; $T_2 \approx 175.7\,N$.

Questions & Discussion

• Question: Does $c$ (the constant on the right side) always have to be zero to use the vector dot product equation?

• Answer: No. If the line passes through the origin, the dot product equals zero ($c=0$). If the line is shifted (has a y-intercept), the dot product equals a non-zero constant ($n \cdot x = c$).

• Question: In vector form ($x = p + su + tv$), must vectors $u$ and $v$ be perpendicular?

• Answer: No, they do not have to be perpendicular, but they must not be parallel to each other. They must intersect/span the plane.

• Course Logistics: The first exam is tentatively scheduled for next Wednesday, covering through section 2.4. Attendance was taken via a digital check-in system.