Study Notes on Vector Calculus and Analytical Geometry - Code 198701008

Core Vector Properties and Operations

• Unit Normal Vectors to a Plane     - A unit normal to the plane defined by two vectors, $\vec{a}$ and $\vec{b}$, is calculated by taking the cross product $\vec{a} \times \vec{b}$ and dividing by its magnitude.     - Given vectors:         - $\vec{a} = \vec{i} + 2\vec{j} + 3\vec{k}$         - $\vec{b} = \vec{i} + \vec{j} + 2\vec{k}$     - Potential solutions provided:         - a) $\vec{i} + \vec{j} - \vec{k}$         - b) $\frac{1}{\sqrt{3}}(\vec{i} + 2\vec{j} - \vec{k})$         - c) $\frac{1}{\sqrt{3}}(\vec{i} + 2\vec{j} + \vec{k})$         - d) $\frac{1}{\sqrt{3}}(\vec{i} + \vec{j} + \vec{k})$         - e) $\frac{1}{\sqrt{3}}(\vec{i} + \vec{j} - \vec{k})$

• Scalar Projection     - The scalar projection of vector $\vec{b}$ on vector $\vec{a}$ is defined as $\text{comp}_{\vec{a}}\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$.     - Given vectors:         - $\vec{a} = \vec{i} + \vec{j} - \vec{k}$         - $\vec{b} = 2\vec{i} + 4\vec{j} + 3\vec{k}$     - Possible values for the scalar projection:         - a) $\sqrt{3}$         - b) $\frac{1}{\sqrt{3}}$         - c) $-\sqrt{3}$         - d) $-\frac{1}{\sqrt{3}}$         - e) $1$

• Relationships Between Vectors and Angles     - If several vectors sum to zero ($\vec{a} + \vec{b} + \vec{c} = \vec{0}$), the law of cosines for vectors can be applied to find internal angles.     - Specific magnitudes provided:         - $\left|\vec{a}\right| = 3$         - $\left|\vec{b}\right| = 5$         - $\left|\vec{c}\right| = 7$     - The angle between $\vec{a}$ and $\vec{b}$ is queried among options:         - a) $60^{\circ}$         - b) $45^{\circ}$         - c) $30^{\circ}$         - d) $90^{\circ}$         - e) none of these

• Vector Linear Combinations     - Analysis of non-collinear vectors $\vec{a}$ and $\vec{b}$.     - Given equations for vectors $\vec{v}$ and $\vec{w}$ involving variables $x$ and $y$:         - $\vec{v} = x\vec{a} + (y - x)\vec{b}$         - $\vec{w} = (2x + y + 3)\vec{a} - (x + 2y)\vec{b}$     - If the condition $3\vec{v} = \vec{w}$ is met, the values of $x$ and $y$ can be solved by comparing coefficients of $\vec{a}$ and $\vec{b}$.     - Options for solutions:         - a) $x = 10, y = 3$         - b) $x = 15, y = -4$         - c) $x = 5, y = 2$         - d) $x = 4, y = -5$         - e) none of these

Analytical Geometry of Lines and Planes

• Intersection of a Line and a Plane     - A line $\vec{r} = (1 + 2t)\vec{i} + t\vec{j} + (1 - t)\vec{k}$ (where $t \in \mathbb{R}$) intersects a plane defined by $2x + y - z = 7$.     - The intersection point is found by substituting the parametric components ($x=1+2t$, $y=t$, $z=1-t$) into the plane equation and solving for $t$.     - Point options:         - a) $(-3, 1, 0)$         - b) $(1, 7, 0)$         - c) $(3, 1, 0)$         - d) $(3, -1, 0)$         - e) $(-3, -1, 0)$

• Deriving the Equation of a Plane     - A plane passing through the point $(1, -1, 3)$ and perpendicular to the vector $\vec{i} + 3\vec{j} - 4\vec{k}$.     - The standard equation of the plane follows the form $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.     - Formula options:         - a) $x - 3y - 4z = 14$         - b) $x - 3y - 4z = -14$         - c) $x + 3y + 4z = -14$         - d) $x + 3y - 4z = -14$         - e) $-x + 3y - 4z = 14$

• Condition for a Line to Lie within a Plane     - Consider the line $\vec{r}(t) = t(2\vec{i} + c\vec{j} + 3\vec{k})$ and the plane $x + 3y + 5z = 0$.     - For the line to lie entirely within the plane, the dot product of the direction vector of the line and the normal vector of the plane must be zero, and the line must contain a point on the plane (in this case, the origin $(0,0,0)$).     - Values for $c$ to satisfy the condition:         - a) $-13/3$         - b) $-11/5$         - c) $17/3$         - d) $-17/3$         - e) none of these

• Tetrahedron Volume Calculations     - The volume $V$ of a tetrahedron with vertices $A, B, C,$ and $D$ is given by $\frac{1}{6} |(\vec{AB} \times \vec{AC}) \cdot \vec{AD}|$     - Coordinates provided:         - $A = (1, 0, 2)$         - $B = (-1, 2, 2)$         - $C = (1, 1, -3)$         - $D = (4, 0, 3)$     - Volume options:         - a) $16/3$         - b) $1/3$         - c) $5$         - d) $11/3$         - e) $32/3$

Advanced Vector Calculus Operators

• Vector Field Operations     - Cross multiplication of vector sums: $(\vec{v} + \vec{p}) \times \vec{w}$.     - Components:         - $\vec{v} = \vec{i} - 2\vec{j} + 3\vec{k}$         - $\vec{w} = -\vec{i} + \vec{j} + 2\vec{k}$         - $\vec{p} = 2\vec{i} + 2\vec{j} - 3\vec{k}$     - Possible result vectors:         - a) $-7\vec{i} + 10\vec{j} + 3\vec{k}$         - b) $7\vec{i} + 3\vec{k}$         - c) $-6\vec{j} + 3\vec{k}$         - d) $-3$         - e) $6\vec{j} + 3\vec{k}$

• Divergence and Gradient Identities     - Calculation of $\text{div}(\text{grad } r^3)$, where $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$ and $r = |\vec{r}|$.     - Possible identities:         - a) $0$         - b) $9r$         - c) $12r$         - d) $6r$         - e) $3r$

• Combinations of Operators (Divergence, Curl, Gradient)     - Discussion of which operator combinations resulting in meaningful expressions where $\vec{F}$ is a vector field and $f$ is a scalar function:         - a) $\text{div}(\text{div } \vec{F})$ (meaningless as divergence results in a scalar, and you cannot take the divergence of a scalar).         - b) $\text{curl}(\text{div } \vec{F})$ (meaningless for the same reason).         - c) $\text{grad}(\text{grad } \vec{F})$.         - d) $\text{grad}(\text{grad } f)$.         - e) $\text{curl}(\text{curl } \vec{F})$.

• Solenoidal Vector Fields     - A vector field $\vec{F}$ is solenoidal if $\text{div } \vec{F} = 0$.     - Vector field: $\vec{F} = (x + 3y)\vec{i} + (y - 3z)\vec{j} + (y + mz)\vec{k}$.     - The condition for $\vec{F}$ being solenoidal determines the constant $m$:         - a) $0$         - b) $1$         - c) $-1$         - d) $2$         - e) $-2$

• Curl Calculation at a Point     - Determining $\text{curl } \vec{F}$ for $\vec{F} = ze^{2xy}\vec{i} + 2xy\cos(y)\vec{j} + (x + 2y)\vec{k}$.     - Specifically evaluated at point $(2, 0, 3)$.     - Options:         - a) $2\vec{i} - 12\vec{k}$         - b) $-2\vec{i} - 12\vec{k}$         - c) $-3\vec{k}$         - d) $3\vec{i} - 12\vec{k}$         - e) $-3\vec{i} - 12\vec{k}$

Multi-Variable Calculus and Surfaces

• Functions and Domains     - Function: $f(x, y, z) = \frac{\sqrt{x - 1}}{\sqrt{1 - y}} + \ln(4 - x^2 - y^2 - z^2)$.     - Domain constraints:         - The numerator $\sqrt{x - 1}$ requires $x \geq 1$.         - The denominator $\sqrt{1 - y}$ requires 1 - y > 0 (since it is in the denominator, it cannot be zero), hence $y < 1$.         - The natural logarithm $\ln(4 - x^2 - y^2 - z^2)$ requires the argument to be positive: $x^2 + y^2 + z^2 < 4$.     - Domain sets:         - a) $(x, y, z) \in \mathbb{R}^3 : y \neq 1$         - b) $(x, y, z) \in \mathbb{R}^3 : x < 1, -2 < y < 1$ and $x^2 + y^2 + z^2 > 4$         - c) (x, y, z) \in \mathbb{R}^3 : -1 < x < 2 and -1 < y < 0         - d) (x, y, z) \in \mathbb{R}^3 : 1 < x < 2, -2 < y < 1 and $x^2 + y^2 + z^2 \leq 4$         - e) (x, y, z) \in \mathbb{R}^3 : 1 \leq x < 2, -2 < y < 1 and x^2 + y^2 + z^2 < 4

• Tangent Plane to a Surface     - Surface equation: $x^2 + 2y^2 + 3z^2 = 36$.     - Evaluation point $P = (1, 2, 3)$.     - The gradient $\nabla f$ at the point provides the normal vector to the tangent plane.     - Plane equation options:         - a) $2x + 8y + 18z = -72$         - b) $x - 4y - z = 10$         - c) $2x + 8y + 18z = 72$         - d) $x + 4y - z = 10$         - e) $x + 4y + z = 10$

• Directional Derivatives     - Function: $f(x, y, z) = 2x^2 + 3y^2 + z^2$.     - Calculation point: $P = (2, 1, 3)$.     - Direction vector: $\vec{i} - 2\vec{k}$.     - The directional derivative is $\nabla f(P) \cdot \vec{u}$, where $\vec{u}$ is the unit vector in the specified direction.     - Result options:         - a) $-\frac{2}{\sqrt{5}}$         - b) $\frac{4}{\sqrt{5}}$         - c) $\frac{2}{\sqrt{5}}$         - d) $\frac{20}{\sqrt{5}}$         - e) $-\frac{4}{\sqrt{5}}$

Kinematics and Curve Properties

• Acceleration Components     - Particle position function: $\vec{r}(t) = \cos(t)\vec{i} + \sin(t)\vec{j} + \cos^2(t)\vec{k}$.     - Goal: Find the tangential component of acceleration ($a_T$) at time $t = \frac{\pi}{4}$.     - Defined as $a_T = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|}$.     - Options: $0, 1, 2, 3, 4$.

• Osculating Plane     - Curve definition: $\vec{r}(t) = (t - \frac{3}{2}\sin(t))\vec{i} + (1 - \frac{3}{2}\cos(t))\vec{j} + t\vec{k}$.     - Point of interest: $(\pi, \frac{5}{2}, \pi)$.     - The osculating plane contains the velocity vector $\vec{r}'(t)$ and the acceleration vector $\vec{r}''(t)$.     - Cartesian equation options:         - a) $2x - 5z = -3\pi$         - b) $2y + z = 1\pi$         - c) $2x + 5z = 3\pi$         - d) $2x + 5z = 1\pi$         - e) $2x - 5z = 3\pi$

• Line Integrals and Arc Length     - Conservative Fields:         - $\vec{F} = (yz)\vec{i} + (zx)\vec{j} + (xy)\vec{k}$.         - Line integral result for any path joining $A = (1, 1, 1)$ and $B = (-1, 2, 0)$.         - Integral output options: $1, 0, 5, -5, -1$.     - Arc Length Formula:         - For curve $\vec{r}(t) = t\vec{i} + t^2\vec{j} + t^3\vec{k}$ for $0 \leq t \leq 1$.         - Formula options:             - a) $\int_0^1 \sqrt{t^2 + t^4 + t^6} dt$             - b) $\int_0^1 \sqrt{1 + 4t^2 + 9t^4} dt$             - c) $\int_0^1 \sqrt{1 + 2t^2 + 3t^6} dt$             - d) $\int_0^1 \sqrt{t + t^2 + t^6} dt$             - e) $\int_0^1 \sqrt{1 + 2t + 3t^2} dt$

Descriptive Geometry and Vector Fields

• Analysis of Plane Alpha ($\alpha$)     - Vector equation: $\vec{r} = (3\vec{i} + 3\vec{j} + 2\vec{k}) + \lambda(-\vec{i} + 2\vec{j} + \vec{k}) + \mu(2\vec{i} + \vec{k})$.     - Perpendicularity check: Show that vector $2\vec{i} + 3\vec{j} - 4\vec{k}$ is perpendicular to plane $\alpha$ (calculated via the cross product of direction vectors $\vec{d_1} \times \vec{d_2}$).     - Cartesian form derivation of plane $\alpha$.     - Shortest distance application:         - Line $l: \vec{r} = (4\vec{i} - 5\vec{j} + 2\vec{k}) + t(\vec{i} + 6\vec{j} - 3\vec{k})$.         - Point $A$ lies on line $l$.         - The shortest distance from $A$ to $\alpha$ is exactly $2\sqrt{29}$.         - Task involves finding possible coordinates of $A$.

• Work and Flux Theorems     - Work Done:         - Force field $\vec{F}(x, y, z) = y^2\vec{i} + (3x - 6y)\vec{j}$.         - Displacement: Line segment from $(3, 7)$ to $(0, 12)$.         - Calculated as $W = \int_C \vec{F} \cdot d\vec{r}$.     - Divergence Theorem Verification:         - Field $\vec{F}(x, y, z) = (x^3 - yz)\vec{i} - 2x^2y\vec{j} + z\vec{k}$.         - Volume: Cube bounded by planes $x = y = z = a$ and the coordinate planes ($x=0, y=0, z=0$).         - Requires computing the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ and comparing it to the volume integral $\iiint_V \text{div } \vec{F} dV$.