 maths > vector-algebra

Types of Vectors

what you'll learn...

Overview

Classification of Vectors

»  Classification by magnitude

→  null or zero vector : magnitude $0$$0$

→  unit vector : magnitude $1$$1$

→  proper vector : magnitude not $0$$0$

»  By similarities of two vectors

→  equal vectors : All corresponding components equal

→  like vectors: same directions

→  unlike vectors : different directions

→  co-initial vectors : same initial point

→  co-linear vectors : on the same line

→  co-planar vectors : on the same plane

→  non-co-planar vectors : not on the same plane

null

The word 'null' means 'zero; nothing'.

A vector with zero magnitude is a 'zero or null vector'.

All the following vectors have $0$$0$ magnitude

$0i+0j+0k$$0 i + 0 j + 0 k$

$0i+0j$$0 i + 0 j$

$0$$0$

A null vector is a scalar as well as a vector.

A null vector is also called improper vector as it does not have a direction.

•  Zero or Null Vector (magnitude $0$$0$)

Null Vector or Zero Vector: A quantity of zero magnitude, given as $\stackrel{\to }{0}$$\vec{0}$ or $0$$0$. For calculations, it can be used as $0i+0j+0k$$0 i + 0 j + 0 k$.

Technically a null vector is not a vector. Arithmetic operations on vectors like addition, may result in a null vector.

$0i$$0 i$ can be called any one of the following:

null vector

improper vector

zero vector

scalar $0$$0$

proper

The word 'proper' means 'correct type or form'.

'A vector with direction' is a 'proper vector'.

Proper Vector: A vector with non-zero magnitude.
If vector $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$ is a proper vector then $\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\ne 0$$\sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \ne 0$

unit

The word 'unit' means $1$$1$

'A vector having magnitude $1$$1$' is an 'unit vector'.

Unit Vector: A vector with magnitude 1.
If vector $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$ is an unit vector, then
$\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}=1$$\sqrt{{a}^{2} + {b}^{2} + {c}^{2}} = 1$

Is $i+j-k$$i + j - k$ an unit vector?

The magnitude is not $1+1-1=1$$1 + 1 - 1 = 1$

The magnitude is $\sqrt{1+1+1}\ne 1$$\sqrt{1 + 1 + 1} \ne 1$

So the given vector is not a unit vactor.

equal

The word 'Equal' means 'being same in quantity or value'.

Two vectors are 'equal' when both the magnitudes and directions are same.

Equal Vectors: The two vectors $\stackrel{\to }{p}={p}_{x}i+{p}_{y}j+{p}_{z}k$$\vec{p} = {p}_{x} i + {p}_{y} j + {p}_{z} k$ and $\stackrel{\to }{q}={q}_{x}i+{q}_{y}j+{q}_{z}k$$\vec{q} = {q}_{x} i + {q}_{y} j + {q}_{z} k$ are Equal
$\stackrel{\to }{p}=\stackrel{\to }{q}$$\vec{p} = \vec{q}$
if and only if
${p}_{x}={q}_{x}$${p}_{x} = {q}_{x}$
${p}_{y}={q}_{y}$${p}_{y} = {q}_{y}$
${p}_{z}={q}_{z}$${p}_{z} = {q}_{z}$

Two vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are equal, then the following are true

$\frac{\stackrel{\to }{p}}{|p|}=\frac{\stackrel{\to }{q}}{|q|}$$\frac{\vec{p}}{| p |} = \frac{\vec{q}}{| q |}$

directional cosines of $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are equal

$|p|=|q|$$| p | = | q |$

like & unlike

The word 'like' means 'having same characteristics or properties'.

Two vectors called 'like vectors' 'when the vectors have same direction'.

Like Vectors: Vectors of same direction.
The vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are like vectors, if
$\frac{\stackrel{\to }{p}}{|p|}=\frac{\stackrel{\to }{q}}{|q|}$$\frac{\vec{p}}{| p |} = \frac{\vec{q}}{| q |}$

Unlike Vectors: Vectors of different direction.
The vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are unlike vectors, if
$\frac{\stackrel{\to }{p}}{|p|}\ne \frac{\stackrel{\to }{q}}{|q|}$$\frac{\vec{p}}{| p |} \ne \frac{\vec{q}}{| q |}$

Two vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are like vectors, then the following are true

$\frac{\stackrel{\to }{p}}{|p|}=\frac{\stackrel{\to }{q}}{|q|}$$\frac{\vec{p}}{| p |} = \frac{\vec{q}}{| q |}$

directional cosines of $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are equal

angles made with $x$$x$, $y$$y$, $z$$z$ axes of $\stackrel{\to }{p}$$\vec{p}$ are equal to that of $\stackrel{\to }{q}$$\vec{q}$.

initial

The word 'initial' means 'beginning or starting'.

The prefix 'co' in co-initial means 'jointly; mutually'.

Two vectors are 'co-initial' vectors 'When the vectors start from the same position'.

A vector can be positioned at any point without modifying the defining parameters magnitude and direction. When vectors are used to represent shapes or quantities, the position of the vector is additionally specified.

Co-initial Vectors:Two vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are co-initial vectors when they are positioned at the same starting point $\left(x,y,z\right)$$\left(x , y , z\right)$.

Given two vectors, $2i+3j$$2 i + 3 j$ and $4i+6j$$4 i + 6 j$, are these vectors co-initial?

The answer is 'Cannot determine from the given information'. The initial position of the vector is to be given separately and when not given, the vectors can be positioned anywhere.

collinear

The word 'collinear' means 'of lying in the same line'. "co" means "together; jointly" ; and "linear" means "line".

Two vectors are called 'collinear' vectors 'When the vectors are on the same line'.

Collinear Vectors: Two vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are collinear vectors if $\stackrel{\to }{p}=n\stackrel{\to }{q}$$\vec{p} = n \vec{q}$ where $n\in ℝ$$n \in \mathbb{R}$.

Given that two vectors are collinear, does that also imply that the given vectors are like vectors?

'No.'

Collinear vectors can be either in same direction or in opposite direction.
Furthermore, there are two parameters to note in collinear vectors when comparing them for being like vectors.

1. the direction - whether they are in same direction or in the opposite direction.

2. the position - vectors may be positioned at different points. Like vectors having same direction may not be collinear because of the position.

co-planar

The word 'co-planar' means 'of lying in the same plane'.

Two vectors are 'coplanar' 'When two vectors are in the same plane'

Co-planar Vectors: Two vectors $\stackrel{\to }{p}$$\vec{p}$ and $\stackrel{\to }{q}$$\vec{q}$ are coplanar if they lie on the same plane.
Under condition that the positions of vectors are not specified, and the vectors can be equivalently placed anywhere in the 3-D space, any two vectors will be coplanar.
Three vectors $\stackrel{\to }{p}$$\vec{p}$, $\stackrel{\to }{q}$$\vec{q}$, $\stackrel{\to }{r}$$\vec{r}$ are co-planar (under the condition that vectors are equivalently positioned anywhere in the 3-D space), if
$|\begin{array}{ccc}{p}_{x}& {p}_{y}& {p}_{z}\\ {q}_{x}& {q}_{y}& {q}_{z}\\ {r}_{x}& {r}_{y}& {r}_{z}\end{array}|=0$$| \left({p}_{x} , {p}_{y} , {p}_{z}\right) , \left({q}_{x} , {q}_{y} , {q}_{z}\right) , \left({r}_{x} , {r}_{y} , {r}_{z}\right) | = 0$
Co-planar property in terms of vector product is given as $\stackrel{\to }{p}\cdot \left(\stackrel{\to }{q}×\stackrel{\to }{r}\right)=0$$\vec{p} \cdot \left(\vec{q} \times \vec{r}\right) = 0$.

Non-co-planar Vectors: Three vectors $\stackrel{\to }{p}$$\vec{p}$, $\stackrel{\to }{q}$$\vec{q}$, $\stackrel{\to }{r}$$\vec{r}$ are non-co-planar (under the condition that vectors are equivalently positioned anywhere in the 3-D space), if
$|\begin{array}{ccc}{p}_{x}& {p}_{y}& {p}_{z}\\ {q}_{x}& {q}_{y}& {q}_{z}\\ {r}_{x}& {r}_{y}& {r}_{z}\end{array}|\ne 0$$| \left({p}_{x} , {p}_{y} , {p}_{z}\right) , \left({q}_{x} , {q}_{y} , {q}_{z}\right) , \left({r}_{x} , {r}_{y} , {r}_{z}\right) | \ne 0$
Co-planar property in terms of vector product is given as $\stackrel{\to }{p}\cdot \left(\stackrel{\to }{q}×\stackrel{\to }{r}\right)\ne 0$$\vec{p} \cdot \left(\vec{q} \times \vec{r}\right) \ne 0$.

Are $3i+4j$$3 i + 4 j$ and $4i-2j$$4 i - 2 j$ coplanar?

The answer is 'Yes, they lie in the xy-plane'.

negative

The word 'negative' means 'opposite; reverse'

The negative of $\stackrel{\to }{p}=ai+bj$$\vec{p} = a i + b j$ is $-\stackrel{\to }{p}=-ai-bj$$- \vec{p} = - a i - b j$

Negative of a Vectors: For the vector $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$, the negative of $\stackrel{\to }{p}$$\vec{p}$ is
$-\stackrel{\to }{p}=-ai-bj-ck$$- \vec{p} = - a i - b j - c k$

What is the 'negative' of vector $\stackrel{\to }{p}=2i-j$$\vec{p} = 2 i - j$?

The answer is '$-2i+j$$- 2 i + j$'.

component

The word 'component' means 'constituent part of a larger whole'.

The components of a vector are 'x, y, and z components along the three axes'.

Component Form of a Vector: A vector $\stackrel{\to }{p}$$\vec{p}$ is given in the component form as
$\stackrel{\to }{p}={p}_{x}i+{p}_{y}j+{p}_{z}k$$\vec{p} = {p}_{x} i + {p}_{y} j + {p}_{z} k$,
where ${p}_{x},{p}_{y},{p}_{z}$${p}_{x} , {p}_{y} , {p}_{z}$ are the components along $x,y,z$$x , y , z$ -axes respectively.

What is the component form of unit vector along $x$$x$-axis?

The answer is '$i$$i$', in the usual convention of representing component along x-axis using $i$$i$.

summary

Classification of Vectors

»  Classification by magnitude

→  null or zero vector : magnitude $0$$0$

→  unit vector : magnitude $1$$1$

→  proper vector : magnitude not $0$$0$<

»  By similarities of two vectors

→  equal vectors : All corresponding components equal

→  like vectors: same directions

→  unlike vectors : different directions

→  co-initial vectors : same initial point

→  co-linear vectors : on the same line

→  co-planar vectors : on the same plane

→  non-co-planar vectors : not on the same plane

Outline