Vector Algebra • Topic 1 of 3

Vectors: Types & Basic Operations

A vector is a quantity with both magnitude and direction — displacement, velocity, force and acceleration are vectors. A scalar has magnitude only (mass, length, temperature, speed). A vector is drawn as a directed line segment $\overrightarrow{AB}$: $A$ is the initial point (tail), $B$ the terminal point (head), and the arrow fixes the direction. The length $|\overrightarrow{AB}|$ is the magnitude.

Component form, magnitude and the unit vector

In space, fixing the standard unit vectors $\hat{i},\hat{j},\hat{k}$ along the axes, any vector is written in component form $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, where $a_1,a_2,a_3$ are its scalar components. Its magnitude follows from Pythagoras in three dimensions:

$$|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$$

A unit vector has magnitude $1$; the unit vector along $\vec{a}$ (its direction, stripped of length) is

$$\hat{a}=\frac{\vec{a}}{|\vec{a}|}\quad(\vec{a}\ne\vec{0}).$$

To build a vector of any required magnitude $m$ in the direction of $\vec{a}$, scale the unit vector: $m\,\hat{a}=\dfrac{m}{|\vec{a}|}\vec{a}$.

Types of vectors

  • Zero (null) vector $\vec{0}$: magnitude $0$; its initial and terminal points coincide, so it has no definite direction. It is the only vector with magnitude $0$.
  • Unit vector: magnitude $1$.
  • Coinitial vectors: share the same initial point.
  • Collinear (parallel) vectors: lie along the same or parallel lines; equivalently $\vec{a}=\lambda\vec{b}$ for some scalar $\lambda$ ($\lambda>0$ same direction, $\lambda<0$ opposite).
  • Equal vectors $\vec{a}=\vec{b}$: same magnitude and same direction, regardless of initial point — equivalently equal components.
  • Negative of a vector: $-\vec{a}$ has the same magnitude as $\vec{a}$ but opposite direction; $\overrightarrow{BA}=-\overrightarrow{AB}$.
  • Position vector: of a point $P(x,y,z)$ relative to the origin $O$, it is $\overrightarrow{OP}=x\hat{i}+y\hat{j}+z\hat{k}$, with $|\overrightarrow{OP}|=\sqrt{x^2+y^2+z^2}$.

Direction cosines and direction ratios

If $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ makes angles $\alpha,\beta,\gamma$ with the positive $x,y,z$ axes, its direction cosines are $l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$, computed as

$$l=\frac{a_1}{|\vec{a}|},\quad m=\frac{a_2}{|\vec{a}|},\quad n=\frac{a_3}{|\vec{a}|},\qquad l^2+m^2+n^2=1.$$

The components $a_1,a_2,a_3$ themselves (or any numbers proportional to $l,m,n$) are called direction ratios. Direction cosines are unique for a given direction; direction ratios are not (any non-zero multiple works). Notice that the direction cosines are exactly the components of the unit vector $\hat{a}$, so $\hat{a}=l\hat{i}+m\hat{j}+n\hat{k}$.

Vector algebra: addition and scalar multiplication

Vectors add componentwise: $(a_1\hat i+a_2\hat j+a_3\hat k)+(b_1\hat i+b_2\hat j+b_3\hat k)=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k$. Geometrically this is the triangle law (place the tail of $\vec{b}$ at the head of $\vec{a}$; the closing side is $\vec{a}+\vec{b}$) or, equivalently, the parallelogram law (place $\vec{a},\vec{b}$ coinitial; the diagonal is the sum). Addition is commutative and associative, with $\vec{a}+\vec{0}=\vec{a}$ and $\vec{a}+(-\vec{a})=\vec{0}$. Multiplying by a scalar $\lambda$ scales each component: $\lambda\vec{a}=\lambda a_1\hat i+\lambda a_2\hat j+\lambda a_3\hat k$, with $|\lambda\vec{a}|=|\lambda|\,|\vec{a}|$.

Vector joining two points and the section formula

If $A,B$ have position vectors $\vec{a},\vec{b}$, the vector from $A$ to $B$ is the head minus the tail:

$$\overrightarrow{AB}=\vec{b}-\vec{a},\qquad |\overrightarrow{AB}|=\text{distance }AB.$$

The point $R$ dividing $AB$ in the ratio $m:n$ has position vector given by the section formula:

$$\text{Internal: }\ \vec{r}=\frac{m\vec{b}+n\vec{a}}{m+n}\qquad\qquad \text{External: }\ \vec{r}=\frac{m\vec{b}-n\vec{a}}{m-n}.$$

The midpoint is the special internal case $m=n=1$, giving $\vec{r}=\dfrac{\vec{a}+\vec{b}}{2}$.

Deeper Insight — one object, two faithful descriptions. The power of vectors is that the geometric picture (an arrow you can slide around) and the algebraic description (a triple of components) are completely interchangeable. Equal vectors may sit anywhere in space yet share one component triple, which is exactly why a force diagram and a coordinate calculation give the same answer. Most early mistakes come from forgetting this: writing $\overrightarrow{AB}=\vec{a}-\vec{b}$ (tail minus head) instead of $\vec{b}-\vec{a}$, or treating direction ratios as if they were unique. Keep "head minus tail" and "components $=$ magnitude $\times$ direction cosines" in mind and the whole chapter stays consistent.

Two vectors a and b drawn as arrows from a common initial point O Coinitial vectors from a common point O O a b c Each arrow's length is its magnitude; the arrowhead fixes direction Triangle law and parallelogram law of vector addition giving the resultant a plus b Triangle law & parallelogram law of addition a b b a + b O A C B Triangle law: head-to-tail. Diagonal of the parallelogram = a + b
1
Worked Example
Find $|\vec a|$ for $\vec a=3\hat i+4\hat k$.
Solution

$|\vec a|=\sqrt{3^2+0^2+4^2}=\sqrt{25}=5.$

Answer: $|\vec a|=5$

2
Worked Example
Find the unit vector along $\vec a=2\hat i+3\hat j+6\hat k$.
Solution

$|\vec a|=\sqrt{4+9+36}=\sqrt{49}=7$. So $\hat a=\dfrac{1}{7}(2\hat i+3\hat j+6\hat k).$

Answer: $\hat a=\dfrac{1}{7}(2\hat i+3\hat j+6\hat k)$

3
Worked Example
Find $\overrightarrow{AB}$ where $A(1,2,3)$, $B(4,6,3)$.
Solution

$\overrightarrow{AB}=\vec b-\vec a=(4-1)\hat i+(6-2)\hat j+(3-3)\hat k=3\hat i+4\hat j.$

Answer: $\overrightarrow{AB}=3\hat i+4\hat j$

4
Worked Example
Find the midpoint of $A(2,0,4)$ and $B(0,2,0)$ as a position vector.
Solution

Midpoint $=\dfrac{\vec a+\vec b}{2}=\dfrac{(2\hat i+4\hat k)+(2\hat j)}{2}=\hat i+\hat j+2\hat k.$

Answer: $\hat i+\hat j+2\hat k$

5
Worked Example
Find a vector of magnitude $15$ in the direction of $\vec a=\hat i-2\hat j+2\hat k$.
Solution

$|\vec a|=\sqrt{1+4+4}=3$, so $\hat a=\dfrac13(\hat i-2\hat j+2\hat k)$. The required vector is $15\hat a=\dfrac{15}{3}(\hat i-2\hat j+2\hat k)=5(\hat i-2\hat j+2\hat k)=5\hat i-10\hat j+10\hat k.$

Answer: $5\hat i-10\hat j+10\hat k$

6
Worked Example
Find the direction cosines of $\vec a=\hat i+2\hat j+2\hat k$.
Solution

$|\vec a|=\sqrt{1+4+4}=3$. The direction cosines are the components of $\hat a$: $l=\dfrac13,\ m=\dfrac23,\ n=\dfrac23$. Check: $l^2+m^2+n^2=\dfrac{1+4+4}{9}=1.$

Answer: $l=\dfrac13,\ m=\dfrac23,\ n=\dfrac23$

7
Worked Example
Show that the points $A(1,2,3),\ B(2,3,4),\ C(3,4,5)$ are collinear.
Solution

$\overrightarrow{AB}=\vec b-\vec a=\hat i+\hat j+\hat k$ and $\overrightarrow{BC}=\vec c-\vec b=\hat i+\hat j+\hat k$. Since $\overrightarrow{BC}=1\cdot\overrightarrow{AB}$, the vectors are parallel and share point $B$, so $A,B,C$ are collinear.

Answer: $A,B,C$ are collinear (since $\overrightarrow{BC}=\overrightarrow{AB}$).

8
Worked Example
Find the position vector of the point $R$ dividing the join of $A(1,2,3)$ and $B(4,5,6)$ in the ratio $2:1$ internally.
Solution

By the internal section formula, $\vec r=\dfrac{2\vec b+1\cdot\vec a}{2+1}=\dfrac{2(4\hat i+5\hat j+6\hat k)+(\hat i+2\hat j+3\hat k)}{3}=\dfrac{9\hat i+12\hat j+15\hat k}{3}=3\hat i+4\hat j+5\hat k.$ So $R=(3,4,5).$

Answer: $R=(3,4,5)$

9
Worked Example
Find a unit vector in the direction of the sum of $\vec a=\hat i+2\hat j-\hat k$ and $\vec b=2\hat i+\hat j+\hat k$.
Solution

$\vec a+\vec b=3\hat i+3\hat j+0\hat k=3(\hat i+\hat j)$, so $|\vec a+\vec b|=3\sqrt{1+1}=3\sqrt2$. The unit vector is $\dfrac{3(\hat i+\hat j)}{3\sqrt2}=\dfrac{1}{\sqrt2}(\hat i+\hat j).$

Answer: $\dfrac{1}{\sqrt2}(\hat i+\hat j)$

10
Worked Example
The point $C$ divides $A(3,-2,1)$ and $B(-1,2,5)$ externally in the ratio $3:1$. Find its position vector.
Solution

By the external section formula with $m=3,n=1$: $\vec r=\dfrac{3\vec b-1\cdot\vec a}{3-1}=\dfrac{3(-\hat i+2\hat j+5\hat k)-(3\hat i-2\hat j+\hat k)}{2}=\dfrac{-6\hat i+8\hat j+14\hat k}{2}=-3\hat i+4\hat j+7\hat k.$ So $C=(-3,4,7).$

Answer: $C=(-3,4,7)$

11
Worked Example
If $\vec a=2\hat i-\hat j+2\hat k$, find $|\vec a|$, $\hat a$ and its direction cosines.
Solution

$|\vec a|=\sqrt{4+1+4}=3$. So $\hat a=\dfrac13(2\hat i-\hat j+2\hat k)$ and the direction cosines are $l=\dfrac23,\ m=-\dfrac13,\ n=\dfrac23$, satisfying $l^2+m^2+n^2=\dfrac{4+1+4}{9}=1.$

Answer: $|\vec a|=3,\ \hat a=\dfrac13(2\hat i-\hat j+2\hat k),\ l=\dfrac23,\ m=-\dfrac13,\ n=\dfrac23$

12
Worked Example
Show that $\vec a=2\hat i+3\hat j$ and $\vec b=4\hat i+6\hat j$ are collinear, and find $\lambda$ with $\vec b=\lambda\vec a$.
Solution

Each component of $\vec b$ is twice the corresponding component of $\vec a$, so $\vec b=2\vec a$ and $\lambda=2$. Equal ratios $\dfrac{4}{2}=\dfrac{6}{3}=2$ confirm the vectors are parallel (collinear).

Answer: $\vec b=2\vec a$, so $\lambda=2$ (the vectors are collinear).

Key Points

  • A vector has magnitude and direction; in components $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ with $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$.
  • Unit vector $\hat a=\dfrac{\vec a}{|\vec a|}$; a vector of magnitude $m$ along $\vec a$ is $m\hat a$.
  • Know the types: zero, unit, coinitial, collinear/parallel ($\vec a=\lambda\vec b$), equal, negative, and position vector $\overrightarrow{OP}$.
  • Direction cosines $l=\dfrac{a_1}{|\vec a|},\,m=\dfrac{a_2}{|\vec a|},\,n=\dfrac{a_3}{|\vec a|}$ satisfy $l^2+m^2+n^2=1$; the components are direction ratios (not unique).
  • Add componentwise (triangle / parallelogram law); $\lambda\vec a$ scales each component and $|\lambda\vec a|=|\lambda||\vec a|$.
  • Vector joining two points: $\overrightarrow{AB}=\vec b-\vec a$ (head minus tail).
  • Section formula — internal $\dfrac{m\vec b+n\vec a}{m+n}$, external $\dfrac{m\vec b-n\vec a}{m-n}$; midpoint $\dfrac{\vec a+\vec b}{2}$.
  • Points are collinear when the joining vectors are parallel (one is a scalar multiple of the other).
Tap an option to check your answer0 / 4
Q1.$|3\hat i+4\hat j|=$
Explanation: $\sqrt{9+16}=5$.
Q2.The unit vector along $2\hat i+3\hat j+6\hat k$ is $\tfrac1k(2\hat i+3\hat j+6\hat k)$ with $k=$
Explanation: $|\vec a|=\sqrt{49}=7$.
Q3.$\overrightarrow{AB}$ equals:
Explanation: Head minus tail position vectors.
Q4.A unit vector has magnitude:
Explanation: By definition, magnitude $1$.