Grade 10

Vectors

Quantities with both magnitude and direction, added and scaled to describe displacement and force.

IntuitionWhat Is a Vector?

When you tell someone you walked three kilometers, they still do not know where you ended up: three kilometers east reaches a completely different place from three kilometers north. Many quantities in physics and geometry—displacement, velocity, acceleration, and force—carry both a magnitude (how much) and a direction (which way). A vector captures both pieces of information in a single mathematical object, drawn as a directed arrow whose length represents magnitude and whose arrowhead shows direction.

Parallelogram law of vector addition: the two vectors

\mathbf{u} = (a_{11}, a_{21})

and

\mathbf{v} = (a_{12}, a_{22})

span a parallelogram whose diagonal is

\mathbf{u} + \mathbf{v}

SchoolAdding and Scaling Vectors

Definition: Directed Segments, Magnitude, and Equality

A vector with initial point $A$ and terminal point $B$ is written $\overrightarrow{AB}$ , or by a single letter $\vec{u}$ . Its magnitude (or length) is the distance between $A$ and $B$ , denoted $|\overrightarrow{AB}|$ or $|\vec{u}|$ . The zero vector $\vec{0}=\overrightarrow{AA}$ has length $0$ and arbitrary direction. Two vectors $\vec{u}$ and $\vec{v}$ are equal ( $\vec{u}=\vec{v}$ ) if and only if they have the same magnitude and the same direction, regardless of where their initial points are placed.

\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}, \qquad \overrightarrow{OA} + \overrightarrow{OB} = \overrightarrow{OC} \quad (OACB \text{ parallelogram})

To add two vectors by the triangle rule, place the tail of the second vector at the tip of the first: going from $A$ to $B$ and then from $B$ to $C$ gives the net displacement $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ . Equivalently, by the parallelogram rule, if $\overrightarrow{OA}$ and $\overrightarrow{OB}$ share the same initial point $O$ , their sum $\overrightarrow{OA}+\overrightarrow{OB}$ is the diagonal $\overrightarrow{OC}$ of the parallelogram $OACB$ . Subtracting vectors is adding the opposite vector: $\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}$ .

|k\vec{u}| = |k|\,|\vec{u}|, \qquad \vec{u}\cdot\vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta

Multiplying a vector $\vec{u}$ by a real number $k$ (called scalar multiplication) produces a vector $k\vec{u}$ of length $|k|\,|\vec{u}|$ that points in the same direction as $\vec{u}$ when $k>0$ , in the opposite direction when $k<0$ , and equals $\vec{0}$ when $k=0$ . The dot product (or scalar product) $\vec{u}\cdot\vec{v}=|\vec{u}|\,|\vec{v}|\cos\theta$ , where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$ ( $0^\circ\le\theta\le 180^\circ$ ), combines two vectors into a single real number that measures how strongly they align; in particular, two nonzero vectors are perpendicular ( $\vec{u}\perp\vec{v}$ ) if and only if $\vec{u}\cdot\vec{v}=0$ .

Core vector operations and their geometric meanings
Operation	Formula	Geometric meaning
Triangle rule	$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$	Chaining displacements tip-to-tail
Subtraction rule	$\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}$	Vector pointing from tip $A$ to tip $B$
Midpoint identity	$\overrightarrow{MA}+\overrightarrow{MB}=\vec{0}$	Midpoint $M$ of $AB$ balances the two endpoints
Dot product	$\vec{u}\cdot\vec{v}=\|\vec{u}\|\,\|\vec{v}\|\cos\theta$	Signed projection of one vector onto another

UndergraduateKey Theorems

Theorem: Centroid Vector Identity

Point $G$ is the centroid of $\triangle ABC$ if and only if $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec{0}$ , or equivalently, for every reference point $O$ in the plane, $\overrightarrow{OG}=\dfrac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$ .

Why is it true?

If equal unit masses are placed at the three vertices $A$ , $B$ , and $C$ , the vectors $\overrightarrow{GA}$ , $\overrightarrow{GB}$ , and $\overrightarrow{GC}$ represent the pulls toward each vertex from $G$ ; their sum is $\vec{0}$ precisely when $G$ is the center of mass.

Proof

Let $M$ be the midpoint of side $BC$ . Since $M$ bisects $BC$ , the two vectors $\overrightarrow{MB}$ and $\overrightarrow{MC}$ have equal length and opposite directions, so $\overrightarrow{MB}+\overrightarrow{MC}=\vec{0}$ . By the triangle rule from $G$ through $M$ , we have $\overrightarrow{GB}+\overrightarrow{GC}=(\overrightarrow{GM}+\overrightarrow{MB})+(\overrightarrow{GM}+\overrightarrow{MC})=2\overrightarrow{GM}$ .

By the median property of $\triangle ABC$ , the centroid $G$ lies on the median $AM$ and divides it in a $2:1$ ratio from the vertex, meaning $\overrightarrow{GA}=-2\overrightarrow{GM}$ .

Adding the two relations yields $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=-2\overrightarrow{GM}+2\overrightarrow{GM}=\vec{0}$ . Finally, for any point $O$ , writing $\overrightarrow{GA}=\overrightarrow{OA}-\overrightarrow{OG}$ (and similarly for $B$ and $C$ ) gives $(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})-3\overrightarrow{OG}=\vec{0}$ , or $\overrightarrow{OG}=\dfrac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$ .

Theorem: Dot Product and the Law of Cosines

For any two vectors $\vec{u}=(x_1,y_1)$ and $\vec{v}=(x_2,y_2)$ forming an angle $\theta$ , the geometric dot product $\vec{u}\cdot\vec{v}=|\vec{u}|\,|\vec{v}|\cos\theta$ equals the coordinate expression $x_1x_2+y_1y_2$ .

Why is it true?

Expanding $|\vec{u}-\vec{v}|^2$ in coordinates and comparing it with the Law of Cosines on the triangle formed by $\vec{u}$ , $\vec{v}$ , and $\vec{u}-\vec{v}$ cancels the squared lengths and isolates the cross-term $x_1x_2+y_1y_2$ .

Proof

Place $\vec{u}=\overrightarrow{OA}$ and $\vec{v}=\overrightarrow{OB}$ at the origin $O$ so that $\overrightarrow{AB}=\vec{v}-\vec{u}=(x_2-x_1,y_2-y_1)$ . By the Law of Cosines in $\triangle OAB$ , the squared side length opposite angle $\theta$ is $|\vec{v}-\vec{u}|^2=|\vec{u}|^2+|\vec{v}|^2-2|\vec{u}|\,|\vec{v}|\cos\theta$ .

On the other hand, computing the squared lengths in coordinates gives $|\vec{u}|^2=x_1^2+y_1^2$ , $|\vec{v}|^2=x_2^2+y_2^2$ , and $|\vec{v}-\vec{u}|^2=(x_2-x_1)^2+(y_2-y_1)^2=x_1^2+y_1^2+x_2^2+y_2^2-2(x_1x_2+y_1y_2)$ .

Equating the two expressions for $|\vec{v}-\vec{u}|^2$ and subtracting $x_1^2+y_1^2+x_2^2+y_2^2$ from both sides yields $-2|\vec{u}|\,|\vec{v}|\cos\theta=-2(x_1x_2+y_1y_2)$ . Dividing by $-2$ gives $\vec{u}\cdot\vec{v}=|\vec{u}|\,|\vec{v}|\cos\theta=x_1x_2+y_1y_2$ .

UndergraduateReal-World Applications and Worked Examples

In mechanics, aviation, and computer graphics, vectors are the standard language for combining motions and forces. An airplane flying through a crosswind moves along the vector sum of its airspeed vector and the wind velocity vector; a physics engine computes mechanical work and surface lighting by taking dot products between force, displacement, and normal vectors.

Example: Resultant of Two Tugboat Forces

Two tugboats pull a barge with forces $\vec{F}_1$ of magnitude $|\vec{F}_1|=300$ kN due east and $\vec{F}_2$ of magnitude $|\vec{F}_2|=400$ kN due north. Find the magnitude $|\vec{F}|$ of the resultant force $\vec{F}=\vec{F}_1+\vec{F}_2$ .

Solution

Choose a coordinate system with the positive first axis pointing east and the positive second axis pointing north, so $\vec{F}_1=(300,0)$ and $\vec{F}_2=(0,400)$ .

Adding the two vectors component-wise gives the resultant force $\vec{F}=\vec{F}_1+\vec{F}_2=(300,400)$ . Its magnitude is $|\vec{F}|=\sqrt{300^2+400^2}=\sqrt{90000+160000}=\sqrt{250000}=500$ kN.

Example: Mechanical Work via the Dot Product

A crate is pulled along a horizontal floor by a rope exerting a force $\vec{F}$ of magnitude $|\vec{F}|=50$ N at an angle $\theta=60^\circ$ above the horizontal over a displacement $\vec{d}$ of length $|\vec{d}|=10$ m. Compute the mechanical work $W=\vec{F}\cdot\vec{d}$ .

Solution

By definition of mechanical work as the dot product of force and displacement, we have $W=\vec{F}\cdot\vec{d}=|\vec{F}|\,|\vec{d}|\cos\theta$ .

Substituting $|\vec{F}|=50$ , $|\vec{d}|=10$ , and $\cos 60^\circ=\dfrac{1}{2}$ gives $W=50\times 10\times\dfrac{1}{2}=250$ J. Only the horizontal component $|\vec{F}|\cos 60^\circ=25$ N along $\vec{d}$ performs work.

What is the magnitude $|\vec{u}|$ of the vector $\vec{u}=(6,-8)$ ?

Let $G$ be the centroid of $\triangle ABC$ . If $\overrightarrow{GA}=(2,-1)$ and $\overrightarrow{GB}=(-5,4)$ , what is $\overrightarrow{GC}$ ?

For which value of $m$ are the vectors $\vec{u}=(3,4)$ and $\vec{v}=(m,-6)$ perpendicular?

A delivery drone flies $12$ km due east from $A$ to $B$ and then $5$ km due north from $B$ to $C$ . What is the magnitude $|\overrightarrow{AC}|$ of its net displacement?

References

H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited
Murray R. Spiegel (1959). Schaum's Outline of Vector Analysis