Mathematics

Vectors in the Plane

Concept of a vector, graphical and analytical operations, magnitude, direction, normalization, and study of the dot product including angle calculation and projections.

A vector v⃗\vec{v} is a directed line segment. It is denoted by a coordinate pair (vx,vy)(v_x, v_y) representing its horizontal and vertical displacement, respectively, in the Cartesian plane R2\mathbb{R}^2.

Ex.: v⃗=(3,2)\vec{v} = (3, 2)
Vector in the coordinate plane indicating its horizontal and vertical components
Vector between two points: Given two points A(ax,ay)A(a_x, a_y) and B(bx,by)B(b_x, b_y), the vector AB⃗\vec{AB} is the directed segment from the initial point AA to the terminal point BB. It is given by:
AB⃗=(bx−ax,by−ay)\vec{AB} = (b_x - a_x, b_y - a_y)
Vector drawn from origin point A to destination point B
Magnitude (or modulus): is the length or size of the vector.
∣v⃗∣=vx2+vy2|\vec{v}| = \sqrt{v_x^2 + v_y^2}
Direction angle: is the angle α\alpha formed with the positive x-axis:
α=arctan⁡ vyvx\alpha = \arctan \, \frac{v_y}{v_x}
Representation of the magnitude of a vector and its angle alpha with respect to the horizontal axis
Unit vector: is a vector with a magnitude of 1.

Normalizing a vector u⃗\vec{u} : means finding a unit vector n⃗u\vec{n}_u that has the same direction as u⃗\vec{u} :
n⃗u=(ux∣u⃗∣,uy∣u⃗∣)\vec{n}_u = \left( \frac{u_x}{|\vec{u}|}, \frac{u_y}{|\vec{u}|} \right)
Addition
Given u⃗=(ux,uy)\vec{u} = (u_x, u_y) and v⃗=(vx,vy)\vec{v} = (v_x, v_y). The sum is:
u⃗+v⃗=(ux+vx,uy+vy)\vec{u} + \vec{v} = (u_x + v_x, u_y + v_y)
Example: u⃗=(2,3);v⃗=(2,−1)  ⟹  u⃗+v⃗=(2+2,3−1)=(4,2)\vec{u} = (2,3); \quad \vec{v} = (2,-1) \implies \vec{u} + \vec{v} = (2+2, 3-1) = (4,2)
Graphical addition of vectors using the polygon or parallelogram rule
Scalar Multiplication
Given v⃗=(vx,vy)\vec{v} = (v_x, v_y) and a scalar kk. The product k⋅v⃗k \cdot \vec{v} is:
k⋅v⃗=(k⋅vx,k⋅vy)k \cdot \vec{v} = (k \cdot v_x, k \cdot v_y)
Example: Let v⃗=(2,1);k=2  ⟹  2⋅v⃗=(2⋅2,2⋅1)=(4,2)\vec{v} = (2,1); \quad k = 2 \implies 2 \cdot \vec{v} = (2 \cdot 2, 2 \cdot 1) = (4,2)
Original vector and the same vector multiplied by 2, being twice as long
Algebraic Definition
Given two vectors u⃗=(ux,uy)\vec{u} = (u_x, u_y) and v⃗=(vx,vy)\vec{v} = (v_x, v_y). The dot product u⃗⋅v⃗\vec{u} \cdot \vec{v} is a scalar given by:
u⃗⋅v⃗=ux⋅vx+uy⋅vy\vec{u} \cdot \vec{v} = u_x \cdot v_x + u_y \cdot v_y
Alternatively, it can be calculated using its geometric definition:
u⃗⋅v⃗=∣u⃗∣⋅∣v⃗∣⋅cos⁡α\vec{u} \cdot \vec{v} = |\vec{u}| \cdot |\vec{v}| \cdot \cos \alpha
Where α\alpha is the angle between the two vectors.

Example: u⃗=(2,3);v⃗=(2,−1)  ⟹  u⃗⋅v⃗=2⋅2+3⋅(−1)=1\vec{u} = (2,3); \quad \vec{v} = (2,-1) \implies \vec{u} \cdot \vec{v} = 2 \cdot 2 + 3 \cdot (-1) = 1
Angle between two vectors
Solving for the angle using the geometric definition of the dot product:
α=arccos⁡(u⃗⋅v⃗∣u⃗∣⋅∣v⃗∣)\alpha = \arccos \left( \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| \cdot |\vec{v}|} \right)
Graphical representation of the angle alpha between vectors u and v
Vector Projection
The projection of vector u⃗\vec{u} onto v⃗\vec{v} is a vector in the direction of v⃗\vec{v}, calculated as:
projv⃗ u⃗=u⃗⋅v⃗∣v⃗∣2 v⃗\text{proj}_{\vec{v}} \, \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \, \vec{v}
Orthogonal projection of vector u onto the direction of vector v
Orthogonal (Perpendicular) Vectors
If two vectors u⃗\vec{u} and v⃗\vec{v} are orthogonal (perpendicular), they form an angle α=90∘\alpha = 90^\circ. Since cos⁡90∘=0\cos 90^\circ = 0, their dot product is zero:
u⃗⋅v⃗=0\vec{u} \cdot \vec{v} = 0
Given a vector v⃗=(vx,vy)\vec{v} = (v_x, v_y), we can easily find a perpendicular vector in the form:
u⃗⊥v⃗=(−vy,vx)or(vy,−vx)\vec{u}_{\perp \vec{v}} = (-v_y, v_x) \quad \text{or} \quad (v_y, -v_x)
Parallel Vectors
If two vectors are parallel, they have the same or opposite direction, meaning their coordinates are proportional. It holds that:
u⃗=k⋅v⃗oruyux=vyvx\vec{u} = k \cdot \vec{v} \quad \text{or} \quad \frac{u_y}{u_x} = \frac{v_y}{v_x}
Distance between two points
The distance between points A(ax,ay)A(a_x, a_y) and B(bx,by)B(b_x, b_y) is equal to the magnitude of the vector connecting them:
d(A,B)=(bx−ax)2+(by−ay)2=∣AB⃗∣d(A, B) = \sqrt{(b_x - a_x)^2 + (b_y - a_y)^2} = |\vec{AB}|
Midpoint of a line segment
The coordinates of the midpoint MM of a segment formed by points A(ax,ay)A(a_x, a_y) and B(bx,by)B(b_x, b_y) are given by the average of their coordinates:
M=(ax+bx2,ay+by2)M = \left( \frac{a_x + b_x}{2}, \frac{a_y + b_y}{2} \right)
Point M located exactly in the middle of the segment formed by A and B
Symmetric Point (Reflection of a point across another point)
Given points A(ax,ay)A(a_x, a_y) and B(bx,by)B(b_x, b_y), the point C(x,y)C(x, y), which is the reflection of AA across BB, must satisfy that vector AB⃗\vec{AB} equals vector BC⃗\vec{BC}:
AB⃗=BC⃗  ⟹  (bx−ax,by−ay)=(x−bx,y−by)\vec{AB} = \vec{BC} \implies (b_x - a_x, b_y - a_y) = (x - b_x, y - b_y)
Vectors AB and BC of equal length and direction showing C as a symmetric point
Example: Find the symmetric point of A(2,3)A(2, 3) with respect to B(1,−2)B(1, -2).
Since AB⃗=(1−2,−2−3)=(−1,−5)\vec{AB} = (1-2, -2-3) = (-1, -5) and BC⃗=(x−1,y−(−2))=(x−1,y+2)\vec{BC} = (x-1, y-(-2)) = (x-1, y+2):
(−1,−5)=(x−1,y+2)(-1, -5) = (x - 1, y + 2)
Equating the coordinates, we get x=0x = 0 and y=−7y = -7, so the point is C(0,−7)C(0, -7).