Vectors and Geometry in Calculus: Study Notes

Vectors and Their Operations

Vector Addition and Scalar Multiplication

Vectors are mathematical objects with both magnitude and direction. In calculus and physics, vectors are often represented in component form, such as u = (a, b) or u = ai + bj.

• Vector Addition: To add two vectors, add their corresponding components.

• Scalar Multiplication: To multiply a vector by a scalar, multiply each component by the scalar.

Example: If u = (3, -1) and v = (-3, 4), then u + 2v = (3, -1) + 2(-3, 4) = (3 - 6, -1 + 8) = (-3, 7).

Expressing Vectors in Component Form

Vectors can be written as ai + bj in two dimensions, or ai + bj + ck in three dimensions, where i, j, k are unit vectors along the x, y, and z axes, respectively.

• Position Vector: The vector from point A to point B is AB = (x_2 - x_1)i + (y_2 - y_1)j.

Example: If A = (-1, -7) and B = (4, -14), then AB = (4 - (-1))i + (-14 - (-7))j = 5i - 7j.

Vector Magnitude and Distance

Magnitude of a Vector

The magnitude (or length) of a vector v = ai + bj is given by:

$|v| = \sqrt{a^2 + b^2}$

For three dimensions, v = ai + bj + ck:

$|v| = \sqrt{a^2 + b^2 + c^2}$

Example: If u = -2i - 7j and v = -4i - 21j, then |v - u| = |(-4 + 2)i + (-21 + 7)j| = |-2i - 14j| = \sqrt{(-2)^2 + (-14)^2} = \sqrt{4 + 196} = \sqrt{200} = 10\sqrt{2}.

Distance Between Two Points

The distance between points P_1(x_1, y_1, z_1) and P_2(x_2, y_2, z_2) is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Example: For P_1 = (1, -1, -2) and P_2 = (5, -6, -5):

$d = \sqrt{(5-1)^2 + (-6+1)^2 + (-5+2)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5\sqrt{2}$

Linear Combinations and Vector Equations

Expressing Vectors as Linear Combinations

Any vector in a plane can be written as a linear combination of two non-parallel vectors. If u = a*v + b*w, solve for scalars a and b by equating components.

• Example: If u = 6i + j, v = i + j, w = i - j, solve for a and b such that u = a*v + b*w:

Set up equations for i and j components and solve the system:

i: $6 = a + b$ j: $1 = a - b$

Solving gives $a = \frac{7}{2}$, $b = \frac{5}{2}$.

Applications of Vectors

Resolving Forces into Components

Forces can be resolved into horizontal (i) and vertical (j) components using trigonometry:

$F = F_x i + F_y j$ where $F_x = F \cos \theta$, $F_y = F \sin \theta$

Example: A force of 8 pounds at 30° with the ground:

$F_x = 8 \cos 30^\circ = 8 \times \frac{\sqrt{3}}{2} = 6.928$ $F_y = 8 \sin 30^\circ = 8 \times \frac{1}{2} = 4.000$

So, $F = 6.928i + 4.000j$

Equilibrium and Tension in Cables

When an object is suspended by two cables, the tensions can be found by resolving forces and setting up equilibrium equations.

• Sum of vertical components equals the weight.

• Sum of horizontal components equals zero.

Example: For a 50 lb speaker hanging from cables at 45° and 35°:

Let $T_1$ and $T_2$ be the tensions. Set up equations: $T_1 \sin 45^\circ + T_2 \sin 35^\circ = 50$ $T_1 \cos 45^\circ = T_2 \cos 35^\circ$ Solve for $T_1$ and $T_2$.

Vectors in Three Dimensions

Position Vectors

The position vector from point P to Q in 3D is:

$\vec{PQ} = (x_2 - x_1)i + (y_2 - y_1)j + (z_2 - z_1)k$

Example: If P = (-1, -3, 0), Q = (4, 3, -3): $\vec{PQ} = (4 - (-1))i + (3 - (-3))j + (-3 - 0)k = 5i + 6j - 3k$

Expressing Vectors in ai + bj + ck Form

Given vectors u and v, and a linear combination such as 2u - 6v, substitute and simplify:

Example: If u = (1, 1, 0), v = (3, 0, 1): 2u - 6v = 2(1, 1, 0) - 6(3, 0, 1) = (2, 2, 0) - (18, 0, 6) = (-16, 2, -6) = -16i + 2j - 6k

Equations of Spheres

Standard Equation of a Sphere

The equation of a sphere with center (h, k, l) and radius r is:

$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$

Expanded form: $x^2 + y^2 + z^2 + Dx + Ey + Fz + G = 0$

Example: Center (8, 10, 0), radius 5: $(x - 8)^2 + (y - 10)^2 + z^2 = 25$ Expanded: $x^2 + y^2 + z^2 - 16x - 20y + 64 + 100 = 25$ Simplify as needed.

Finding Center and Radius from General Equation

Given $x^2 + y^2 + z^2 - 18x - 10y - 6z = -15$, complete the square for each variable:

• Group terms: $(x^2 - 18x) + (y^2 - 10y) + (z^2 - 6z) = -15$

• Complete the square for each group:

• $(x^2 - 18x + 81)$, $(y^2 - 10y + 25)$, $(z^2 - 6z + 9)$

• Add 81, 25, and 9 to both sides: $-15 + 81 + 25 + 9 = 100$

• So, $(x - 9)^2 + (y - 5)^2 + (z - 3)^2 = 100$

• Center: (9, 5, 3), Radius: 10

Geometric Descriptions of Sets Defined by Inequalities

Sets of points defined by inequalities involving $x^2 + y^2 + z^2$ describe regions in space:

• $x^2 + y^2 + z^2 > 1$ describes all points outside the sphere of radius 1 centered at the origin.

Summary Table: Key Vector Operations

Operation	Formula	Example
Vector Addition	$(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)$	$(3, -1) + (-3, 4) = (0, 3)$
Scalar Multiplication	$k(a, b) = (ka, kb)$	$2(1, 2) = (2, 4)$
Magnitude	$|v| = \sqrt{a^2 + b^2}$	$|(-2, -14)| = \sqrt{(-2)^2 + (-14)^2} = 10\sqrt{2}$
Distance (3D)	$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$	$(1, -1, -2)$ to $(5, -6, -5)$: $5\sqrt{2}$
Equation of Sphere	$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$	Center (9, 5, 3), radius 10: $(x - 9)^2 + (y - 5)^2 + (z - 3)^2 = 100$