Vectors. Properties, Forces, and Planes.

All things are difficult before they are easy, Thomas Fuller.

When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes.

Recall: Vector Basics

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., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.

Basic Vector Operations

• Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.

• Vector Subtraction: 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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.

• 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. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$

Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$

• The projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.

Solved exercises

• Finding a Unit Vector. A vector $\vec{v}$ has an initial point $\langle -1, -3 \rangle$ and a terminal point $\langle 2, 1 \rangle$. Find the unit vector in the direction of $\vec{v}$.

1. Find the components of the vector:
   We determine the vector $\vec{v}$ by subtracting the coordinates of the initial point from the terminal point: $\vec{v} = \langle 2 - (-1), 1 - (-3) \rangle = \langle 2 + 1, 1 + 3 \rangle = \langle 3, 4 \rangle$

2. Calculate the magnitude (length):
   Using the Pythagorean theorem, the magnitude $\vert{}\vert{}\vec{v}\vert{}\vert{}$ is: $\vert{}\vert{}\vec{v}\vert{}\vert{} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

3. Find the unit vector:
   A unit vector $\vec{u}$ has a length of 1. We find it by scaling $\vec{v}$ down by its magnitude (dividing each component by 5): $\vec{u} = \frac{\vec{v}}{\vert{}\vert{}\vec{v}\vert{}\vert{}} = \langle \frac{3}{5}, \frac{4}{5} \rangle$

• Vector Property Proofs. Prove the commutative property of vector addition and the distributive property of scalar multiplication for 2D vectors.

Let $r$ be any scalar, and let $\vec{u} = \langle x_1, y_1 \rangle$ and $\vec{v} = \langle x_2, y_2 \rangle$ be any two vectors.

Commutative Property of Addition, $\vec{u} + \vec{v} = \vec{v} + \vec{u}$:

1. Vector Addition Definition: $\vec{u} + \vec{v} = \langle x_1 + x_2, y_1 + y_2 \rangle$
2. Commutative Property of Real Numbers: Since scalar addition is commutative, we can swap the terms inside the components: $\langle x_2 + x_1, y_2 + y_1 \rangle$
3. Re-apply Vector Addition Definition: $\langle x_2 + x_1, y_2 + y_1 \rangle = \vec{v} + \vec{u}$

Distributive Property $r(\vec{u} + \vec{v}) = r\vec{u} + r\vec{v}$:

1. Vector Addition Definition: $r(\vec{u} + \vec{v}) = r\langle x_1 + x_2, y_1 + y_2 \rangle$
2. Scalar Multiplication Definition: $= \langle r(x_1 + x_2), r(y_1 + y_2) \rangle$
3. Distributive Property of Real Numbers: $= \langle rx_1 + rx_2, ry_1 + ry_2 \rangle$
4. Separate into two vectors: $= \langle rx_1, ry_1 \rangle + \langle rx_2, ry_2 \rangle$
5. Factor out the scalar: $= r\vec{u} + r\vec{v}$

• Finding the Angle in a 3D Triangle. Points $P = (1, 0, 0)$, $Q = (0, 1, 0)$, and $R = (0, 0, 2)$ form a triangle. Calculate the angle $\theta$ formed by the sides $PQ$ and $PR$ (Figure vi).

1. To find the angle between two vectors, we use the geometric definition of the dot product: $\vec{PQ} \cdot \vec{PR} = \vert{}\vert{}\vec{PQ}\vert{}\vert{} \cdot \vert{}\vert{}\vec{PR}\vert{}\vert{} \cos(\theta)$.
2. Define the vectors forming the sides of the angle at point P:
   $\vec{PQ}$ represents the displacement from $P$ to $Q$: $\vec{PQ} = \langle 0-1, 1-0, 0-0 \rangle = \langle -1, 1, 0 \rangle$
   $\vec{PR}$ represents the displacement from $P$ to $R$: $\vec{PR} = \langle 0-1, 0-0, 2-0 \rangle = \langle -1, 0, 2 \rangle$
3. Calculate the dot product: $\vec{PQ} \cdot \vec{PR} = (-1)(-1) + (1)(0) + (0)(2) = 1 + 0 + 0 = 1$
4. Calculate the magnitudes: $\vert{}\vert{}\vec{PQ}\vert{}\vert{} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}$, $\vert{}\vert{}\vec{PR}\vert{}\vert{} = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{5}$
5. Solve for $\theta$: $1 = \sqrt{2} \cdot \sqrt{5} \cdot \cos(\theta) \implies \cos(\theta) = \frac{1}{\sqrt{10}}$
   $\theta = \arccos\left(\frac{1}{\sqrt{10}}\right) \approx 71.5^\circ$

• Martha’s car is stuck in the mud. A truck pulls it with a horizontal force of 300 lb, while Martha and her friend push at a 15° upward angle with a force of 150 lb. Calculate the resultant force’s magnitude and direction angle from the positive x-axis (Figure A).

1. We must break all forces down into their $x$ (horizontal) and $y$ (vertical) components before we can add them together.
2. The truck’s force can be represented as a horizontal vector, $\vec{F_1}$, with a magnitude of 300 lb.
   Because it pulls perfectly horizontally, all its force is in the $x$-direction, $\vec{F_1} = \langle 300, 0 \rangle$
3. Martha and her friend’s pushing force can be represented as a vector $\vec{F_2}$ with a magnitude of 150 lb at an angle of 15° to the truck’s force.
   We use trigonometry to split the angled force into components and we get: $\vec{F_2} = \langle 150\cos(15^\circ), 150\sin(15^\circ) \rangle \approx \langle 150 \cdot 0.9659, 150 \cdot 0.2588 \rangle \approx \langle 144.9, 38.8 \rangle$
4. Add the vectors to find the resultant force ($\vec{R}$): $\vec{R} = \vec{F_1} + \vec{F_2} = \langle 300 + 144.9, 0 + 38.8 \rangle = \langle 444.9, 38.8 \rangle$
5. Calculate the resultant force Magnitude and Direction:
   Magnitude: $\vert{}\vert{}\vec{R}\vert{}\vert{} = \sqrt{444.9^2 + 38.8^2} = \sqrt{197936.01 + 1505.44} \approx 446.6 \text{ lb}$
   Direction Angle: $\theta = \arctan\left(\frac{38.8}{444.9}\right) \approx \arctan(0.0872) \approx 5^\circ$
6. Conclusion: Hence, the resultant force acting on the car is approximately 447.7 lb, and the direction angle of this force from the positive x-axis is approximately 5°.

• Defining a Plane in 3D Space. What is the set of points defined by the equation $x + 2y + 3z = 0$ (Figure A)?

1. The general standard form for the equation of a plane in 3D space is $Ax + By + Cz = D$. A plane is uniquely defined by two things: (i) A known point on the plane. (ii) A normal vector (a vector perfectly perpendicular to the entire plane).
2. In the equation $1x + 2y + 3z = 0$:
   The coefficients of $x$, $y$, and $z$ always give us the normal vector $\vec{A}$. Therefore, $\vec{A} = \langle 1, 2, 3 \rangle$.
   Because $D = 0$, the origin point $O = (0, 0, 0)$ satisfies the equation ($0 + 0 + 0 = 0$). This means the plane passes exactly through the origin.
3. Understanding it via the Dot Product:
   Let $P = (x, y, z)$ be any generic point on this plane. The position vector from the origin to this point is $\vec{OP} = \langle x, y, z \rangle$.
   Because $\vec{A}$ is perpendicular (orthogonal) to the plane, it must be perpendicular to every vector lying on the plane, including $\vec{OP}$. Therefore, their dot product must equal zero: $\vec{A} \cdot \vec{OP} = 0$
4. In our particular case, $\vec{A} = \langle 1, 2, 3 \rangle$, so $\langle 1, 2, 3 \rangle \cdot \langle x, y, z \rangle = 0$.
   $1x + 2y + 3z = 0$
   This proves that the set of all points $(x,y,z)$ that satisfy this equation form a plane passing through the origin, perpendicular to the vector $\langle 1, 2, 3 \rangle$.

Calculate or express the Area of a Triangle using Vectors (Figure D)

Geometric Foundation

Let $\vec{A}$ and $\vec{B}$ be two vectors forming the sides of a triangle, originating from the same point. The standard geometric area of a triangle is half the base times the height. $\text{Area}_{\text{Triangle}} = \frac{\text{base} \times \text{height}}{2}$.

If we treat the magnitude $\vert{}\vert{}\vec{A}\vert{}\vert{}$ as the base, the height $h$ can be found using the angle $\theta$ between the two vectors: $\sin(\theta) = \frac{h}{\vert{}\vert{}\vec{B}\vert{}\vert{}} \implies h = \vert{}\vert{}\vec{B}\vert{}\vert{} \sin(\theta)$

Substituting this into the standard area formula gives the vector-based expression: $Area_{Triangle} = \frac{\text{base} \times \text{height}}{2} = \frac{1}{2} \cdot \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cdot \sin(\theta)$

2D Vectors & The Determinant

1. To calculate this directly using 2D vector components $\vec{A} = \langle a_1, a_2 \rangle$ and $\vec{B} = \langle b_1, b_2 \rangle$, we can use a geometric trick: rotating $\vec{A}$ by $90^\circ$ to link the dot product to the area.
2. Rotate the Vector: Rotate $\vec{A}$ counter-clockwise by $90^\circ$ to create a new vector $\vec{A'} = \langle -a_2, a_1 \rangle$.
3. The new angle $\theta'$ between $\vec{A'}$ and $\vec{B}$ satisfies: $\theta' = \frac{\pi}{2}-\theta'$. Therefore, $\theta$ and $\theta'$ are complementary angles, $\cos(\theta') = \sin(\theta)$.
4. Take the Dot Product: $\vec{A'} \cdot \vec{B} = \vert{}\vert{}\vec{A'}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cdot \cos(\theta')$
   Since the length hasn’t changed ($\vert{}\vert{}\vec{A'}\vert{}\vert{} = \vert{}\vert{}\vec{A}\vert{}\vert{}$) and $\cos(\theta') = \sin(\theta)$, this perfectly matches our previously calculated area formula (without the $\frac{1}{2}$): $\vec{A'} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cdot \sin(\theta)$
5. Applying the standard dot product formula to the components: $\vec{A'} \cdot \vec{B} = \langle -a_2, a_1 \rangle \cdot \langle b_1, b_2 \rangle = a_1b_2 - a_2b_1$
   The result $a_1b_2 - a_2b_1$ is the determinant of the matrix formed by the two vectors. It gives the signed area of the full parallelogram formed by $\vec{A}$ and $\vec{B}$ (Figure ii). We halve it to find the triangle’s area: $\boxed{Area_{Triangle} = \frac{1}{2} \vert{} a_1b_2 - a_2b_1 \vert{} = \frac{1}{2} \left\vert{} \det \begin{pmatrix} a_1 & a_2 \\ b_1 & b_2 \end{pmatrix} \right\vert{}}$

3D Vectors & The Cross Product

For vectors in 3D space, the cross product $\vec{A} \times \vec{B}$ yields a new orthogonal vector whose magnitude equals the area of the spanned parallelogram.

• Find the area of the triangle with vertices $A(1, 2, 3)$, $B(2, -1, 4)$, and $C(4, 5, -1)$.

1. Define the side vectors.
   $\vec{AB} = \langle 2-1, -1-2, 4-3 \rangle = \langle 1, -3, 1 \rangle$
   $\vec{AC} = \langle 4-1, 5-2, -1-3 \rangle = \langle 3, 3, -4 \rangle$
2. Compute the cross product.
   $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$
   $= \hat{\mathbf{i}}((-3)(-4) - (1)(3)) - \hat{\mathbf{j}}((1)(-4) - (1)(3)) + \hat{\mathbf{k}}((1)(3) - (-3)(3)) = \hat{\mathbf{i}}(12 - 3) - \hat{\mathbf{j}}(-4 - 3) + \hat{\mathbf{k}}(3 + 9) = \langle 9, 7, 12 \rangle$
3. Calculate the Area of the triangle.
   The area is half the magnitude of the cross product: $Area = \frac{1}{2} \vert{}\vert{}\vec{AB} \times \vec{AC}\vert{}\vert{} = \frac{1}{2} \sqrt{9^2 + 7^2 + 12^2} = \frac{1}{2} \sqrt{274} \approx 8.28$

Bibliography

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5. Michael Penn, and MathMajor.
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