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High schoolπŸ“Geometry12 min read
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Vectors: Complete Lesson (coordinates, collinearity, dot product)

Everything on planar vectors: definition, coordinates, addition, scalar multiplication, collinearity, norm, dot product. Grade 10 lesson.

Prerequisites :

  • Coordinate plane
  • Square roots
  • Basic trigonometry

Definition

A vector AB⃗\vec{AB} is determined by three things:

  • A direction (that of line (AB)(AB))
  • A sense (from AA to BB)
  • A magnitude (the length ABAB)

Two vectors with same direction, sense and magnitude are equal, even if they start from different points.

Coordinates in a frame

In a frame (O,i⃗,j⃗)(O, \vec{i}, \vec{j}), if A(xA,yA)A(x_A, y_A) and B(xB,yB)B(x_B, y_B):

ABβƒ—(xBβˆ’xAyBβˆ’yA)\vec{AB} \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}

Watch out: β€œendpoint minus start”, not the other way.

Operations

Addition (Chasles)

AB⃗+BC⃗=AC⃗\vec{AB} + \vec{BC} = \vec{AC}

In coordinates: add component-wise.

Scalar multiplication

If k∈Rk \in \mathbb{R} and uβƒ—=(x,y)\vec{u} = (x, y): kuβƒ—=(kx,ky)k \vec{u} = (kx, ky)

  • k>0k > 0: same sense, norm multiplied by ∣k∣|k|.
  • k<0k < 0: opposite sense.
  • k=0k = 0: zero vector.

Norm of a vector

βˆ₯uβƒ—βˆ₯=x2+y2\|\vec{u}\| = \sqrt{x^2 + y^2}

The vector’s length.

Example: uβƒ—(3,4)\vec{u}(3, 4) β‡’ βˆ₯uβƒ—βˆ₯=5\|\vec{u}\| = 5.

Collinearity

Vectors u⃗(x1,y1)\vec{u}(x_1, y_1) and v⃗(x2,y2)\vec{v}(x_2, y_2) are collinear iff:

det⁑(uβƒ—,vβƒ—)=x1y2βˆ’x2y1=0\det(\vec{u}, \vec{v}) = x_1 y_2 - x_2 y_1 = 0

Uses:

  • 3 points A,B,CA, B, C collinear ⇔ ABβƒ—,ACβƒ—\vec{AB}, \vec{AC} collinear.
  • Parallel lines ⇔ collinear direction vectors.

Dot product

u⃗⋅v⃗=xuxv+yuyv\vec{u} \cdot \vec{v} = x_u x_v + y_u y_v

Also: uβƒ—β‹…vβƒ—=βˆ₯uβƒ—βˆ₯βˆ₯vβƒ—βˆ₯cos⁑(ΞΈ)\vec{u} \cdot \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos(\theta)

Key property (orthogonality)

uβƒ—βŠ₯vβƒ—β€…β€ŠβŸΊβ€…β€Šuβƒ—β‹…vβƒ—=0\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = 0

The perpendicularity test.

Midpoint

If II is the midpoint of [AB][AB]: I(xA+xB2,yA+yB2)I\left(\dfrac{x_A + x_B}{2}, \dfrac{y_A + y_B}{2}\right)

Vectorially: AI⃗=IB⃗\vec{AI} = \vec{IB}.

Sample exercise

A(1,2)A(1, 2), B(4,6)B(4, 6), C(7,10)C(7, 10).

Q1: Are they collinear?

ABβƒ—=(3,4)\vec{AB} = (3, 4), ACβƒ—=(6,8)\vec{AC} = (6, 8). det⁑=3Γ—8βˆ’6Γ—4=0\det = 3 \times 8 - 6 \times 4 = 0. Yes.

Q2: Distance ABAB.

βˆ₯ABβƒ—βˆ₯=25=5\|\vec{AB}\| = \sqrt{25} = 5.

Q3: Midpoint MM of [AC][AC].

M(4,6)M(4, 6). Notice: M=BM = B! (points are collinear).

Common mistakes

  1. Confusing point and vector coordinates: ABβƒ—=(xBβˆ’xA,yBβˆ’yA)\vec{AB} = (x_B - x_A, y_B - y_A).
  2. Using slope to test collinearity: determinant is safer (no division by zero).
  3. Adding norms: βˆ₯uβƒ—+vβƒ—βˆ₯β‰ βˆ₯uβƒ—βˆ₯+βˆ₯vβƒ—βˆ₯\|\vec{u} + \vec{v}\| \neq \|\vec{u}\| + \|\vec{v}\| generally.

Takeaways

  • Vector coordinates: endpoint βˆ’ start.
  • βˆ₯uβƒ—βˆ₯=x2+y2\|\vec{u}\| = \sqrt{x^2 + y^2}.
  • Collinear ⇔ det⁑=0\det = 0.
  • Orthogonal ⇔ dot product = 0.
  • Chasles: ABβƒ—+BCβƒ—=ACβƒ—\vec{AB} + \vec{BC} = \vec{AC}.
  • Midpoint: average of coordinates.

Going further

#vectors#coordinates#collinearity#dot product#grade 10