Basic Concepts of Vector Algebra

INTRODUCTION

A vector may be described as a quantity having both magnitude and direction.

Geometrically a vector is represented by a directed line segment as shown in the adjacent figure. The direction is indicated by the length of the segment AB. A is called the initial point and B the terminal point of vector $\overline{AB}$ = $\bar{a}$.

The line l of which the segment AB is a part, is called the line of support. The length is denoted by |$\overline{AB}$| = |$\bar{a}$| = AB = a.

A scalar is quantity, having magnitude only. In other words, a scalar is just a real number.

Displacement, velocity, momentum, area are some examples of vectors while distance, speed, volume temperature are just scalars.

FREE VECTORS

Vectors which are fully characterized by the magnitude and direction only are called free vectors and those are fully characterized by the magnitude, direction and also line of support are called line (or bound) vectors.

Displacement, velocity are free vectors while force and moment of a force about a point are line vectors.

Free Vector: A free vector is not located in any particular position. If a free vector $\overrightarrow{V}$ can be represented by $\overrightarrow{OP}$, it can equally be represented by $\overrightarrow{AB}$, where OP and AB are equal in length and are in the same direction (i.e. $\overrightarrow{AB}$ = $\overrightarrow{OP}$).

AB = OP = |$\overrightarrow{V}$|

Also AB is in the same direction as OP.

POSITION VECTOR OF A POINT

We take arbitrarily any point O in space to be called the origin of reference. The position vector (p.v.) of any point P, with respect to the origin is the vector $\overline{OP}$. For any two points P and Q in space, the equality $\overline{PQ}$ = $\overline{OQ}-\overline{OP}$ expresses any vector $\overline{PQ}$in terms of the position vectors $\overline{OP}$and $\overline{OQ}$of P and Q respectively.

ANGLE BETWEEN TWO VECTORS

It is defined as the smaller angle formed when the initial points or the terminal points of two vectors are brought together.

Note: $\bullet$ 0° $⩽$ $\theta$ $⩽$ 180°

MULTIPLICATION OF A VECTOR BY A SCALAR

Given a vector$\bar{a}$and a scalar k$\in$R, then k$\bar{a}$ (or $\bar{a}$k) denotes a vector whose magnitude is $|k|$ $|\bar{a}|$i.e., k times that of $\bar{a}$and whose direction is the same or opposite to that of $\bar{a}$according as k > 0 or k < 0 respectively. Also, 0$\bar{a}$ = $\bar{o}$, zero or null vector which has zero magnitude and arbitrary direction.

1$\bar{a}$ = $\bar{a}$, (-1) $\bar{a}$ = -$\bar{a}$, etc.

When we have two vectors $\bar{a}$and $\bar{b}$such that $\bar{b}$= k$\bar{a}$, k$\in$R, then $\bar{a}$and $\bar{b}$are called collinear vectors. $\bar{b}$ is said to be a scalar multiple of $\bar{a}$. $\bar{a}$ and $\bar{b}$are parallel if k > 0 and anti parallel if k < 0.

Note also that k₁ (k_2$\bar{a}$) = (k₁k₂) $\bar{a}$ = k₂(k_1$\bar{a}$) $\forall$ k₁, k_{2 $\in$} R.

A vector having the magnitude as one (unity) is called a unit vector.

Unit vector in the direction of $\bar{a}$ is defined as $\frac{1}{|\bar{a}|}$$\bar{a}$ = $\frac{\bar{a}}{|\bar{a}|}$and is denoted by $\hat{a}$.;ADITTION OF TWO VECTORS;TRIANGLE LAW OF ADDITION
Given two vectors $\bar{a}$ and $\bar{b}$, their sum or resultant written as ($\bar{a}$ + $\bar{b}$) is a vector obtained by first bringing the initial point of $\bar{b}$to the terminal point of $\bar{a}$and then joining the initial point of $\bar{a}$to the terminal point of $\bar{b}$giving a consistent direction by completing the triangle OAB

PARALLELOGRAM LAW OF ADDITION ;The sum can also be obtained by bringing the initial points of $\bar{a}$and $\bar{b}$together and then completing the parallelogram OACB;;Note that addition is commutative i.e., $\bar{a}$+ $\bar{b}$= $\bar{b}+\bar{a}$
Also, $\bar{a}$ + ($\bar{b}$+$\bar{c}$) = ($\bar{a}$ + $\bar{b}$) + $\bar{c}$ i.e. the addition of vectors obeys the associative law. If $\bar{a}$ and $\bar{b}$are collinear, their sum is still obtained in the same manner although we do not have a triangle or a parallelogram in this case.;POLYGON LAW OF ADDITION ;For adding more than two vectors, we have a polygon law of addition which is just an extension of the triangle law.$\overline{OA}+\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}=\overline{OF}$A consequence of this is that, if the terminus of the last vector coincides with the initial point of the first vector, the sum of the vectors is $\bar{O}$. To obtain $\bar{a}-\bar{b}$ (difference of two vectors), perform addition of$\bar{a}$ and $(-\bar{b})$.Also, $\bar{a}+\bar{o}$ = $\bar{a}$; $\bar{a}$ + $(-\bar{a})$ = $\bar{o}$;(k₁ + k₂) $\bar{a}$ = k_1$\bar{a}$ + k_2$\bar{a}$; k ($\bar{a}+\bar{b})$ = k$\bar{a}$ + k$\bar{b}$. ;
PROPERTIES OF VECTOR ADDITION $\bullet$ $\bar{a}+\bar{b}=\bar{b}+\bar{a}$;;;;;;;vector addition is commutative$\bullet$ $\bar{a}+(\bar{b}+\bar{c})=(\bar{a}+\bar{b})+\bar{c}$ vector addition is associative$\bullet$ $\bar{a}+\bar{0}=\bar{a}$$\bullet$ $\bar{a}+(-\bar{a})=\bar{0}$$\bullet$ (k₁ + k₂)$\bar{a}$ = k_1$\bar{a}$ + k₂. $\bullet$ k$(\bar{a}+\bar{b})$ = $k\bar{a}+k\bar{b}$;Illustration 1:;;;;;;;;;;;ABCD is a parallelogram A₁ and B₁ are the midpoints of side BC and CD respectively. If $\overrightarrow{A{A}_1}+\overrightarrow{A{B}_1}=\lambda \overrightarrow{AC}$ then find the value of $\lambda$.Solution:;;;;;;;;;;;;;;;;;;Let P.V. of A, B, D be $\vec{0},\vec{b}and\vec{d}$ respectively. Then P.V. of C = $\vec{b}+\vec{d}$.

Also P.V. of $A_1=\vec{b}+\frac{\vec{d}}{2}$ and P.V. of $B_1=\vec{d}+\frac{\vec{b}}{2}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ $\overrightarrow{A{A}_1}+\overrightarrow{A{B}_1}=\frac{3}{2}(\vec{b}+\vec{d})=\frac{3}{2}\overrightarrow{AC}$, hence the value of $\lambda$ is 3/2.
Fundamental theorems of vectors;FUNDAMENTAL THEOREM OF VECTORS IN TWO-DIMENSIONSIf $\bar{a}$and $\bar{b}$be two non-zero non-collinear vectors, then any vector $\bar{r}$in the plane of $\bar{a}$and $\bar{b}$can be expressed uniquely as a linear combination of $\bar{a}$and $\bar{b}$i.e. there exist unique l, m$\in$R such that l$\bar{a}$ +m$\bar{b}$ = $\bar{r}$.This also means that if;l_1$\bar{a}$ + m_1$\bar{b}$ = l_2$\bar{a}$ + m_2$\bar{b}$ then l₁ = l₂ and m₁ = m₂.;FUNDAMENTAL THEOREM OF VECTORS IN THREE-DIMENSIONSIf $\bar{a}$, $\bar{b}$and $\bar{c}$be three non-zero, non-coplanar vectors in space, then any vector $\bar{r}$in space can be expressed uniquely as a linear combination of $\bar{a}$, $\bar{b}$and $\bar{c}$. i.e there exist unique l, m, n $\in$R such that l$\bar{a}$ + m$\bar{b}$ + n$\bar{c}$ = $\bar{r}$This also means that if l_1$\bar{a}$ + m_1$\bar{b}$+ n_1$\bar{c}$ = l₂ $\bar{a}$+ m_2$\bar{b}$+ n_2$\bar{c}$, then l₁ = l₂, m₁ = m₂ and n₁ = n₂.;LINEAR COMBINATIONS OF VECTORS;The linear combination of a finite set of vectors ${\bar{a}}_1,{\bar{a}}_2$,…${\bar{a}}_n$is defined as a vector $\bar{r}$such that $\bar{r}$ = $k_1{\bar{a}}_1$+ $k_2{\bar{a}}_2$+ ……+ $k_n{\bar{a}}_n$, where k₁, k₂, … k_n are any scalars (real numbers).;LINEARLY DEPENDENT AND INDEPENDENT VECTORS;A system of vectors ${\bar{a}}_1,{\bar{a}}_2,\dots ,{\bar{a}}_n$is said to be linearly dependent if there exists a system of scalars k₁, k₂ …, k_n (not all zero) such that k_{1${\bar{a}}_1$} + k_{2${\bar{a}}_2$}+ … +k_{n${\bar{a}}_n$}= $\bar{O}.$They are said to be linearly independent if every relation of the type k_{1${\bar{a}}_1$} + k_{2${\bar{a}}_2$}+ … +k_{n${\bar{a}}_n$}= $\bar{O}$ implies that k₁ = k₂ =….=k_n = 0.;Notes:;;;;;$\bullet$Two collinear vectors are always linearly dependent.$\bullet$Two non-collinear non-zero vectors are always linearly independent$\bullet$Three coplanar vectors are always linearly dependent.$\bullet$Three non-coplanar non-zero vectors are always linearly independent.$\bullet$ More than 3 vectors are always linearly dependent.$\bullet$ Three points with position vectors $\bar{a},\bar{b},\bar{c}$are collinear if $\lambda$_{1$\bar{a}+{\lambda }_2\bar{b}+{\lambda }_3\bar{c}=0$.} with _$\lambda$1 + $\lambda$₂ + $\lambda$₃ = 0.$\bullet$ Four points with position vectors $\bar{a},\bar{b},\bar{c}$,$\bar{d}$are coplanar if $\lambda$_{1$\bar{a}+{\lambda }_2\bar{b}+{\lambda }_3\bar{c}+{\lambda }_4\bar{d}=0$} with $\lambda$₁ + $\lambda$₂ + $\lambda$₃ + $\lambda$₄= 0.;Illustration 2:;;;;;;;;;;;If $\vec{a},\vec{b},\vec{c}$ are non–zero non coplanar vectors determine whether the vectors:${\vec{r}}_1=2\vec{a}-3\vec{b}+\vec{c},{\vec{r}}_2=3\vec{a}-5\vec{b}+2\vec{c}$ and ${\vec{r}}_3=4\vec{a}-5\vec{b}+\vec{c}$ are linearly independent or dependent. ;Solution:;;;;;;;;;;;;;;;;;;Let ${\vec{r}}_3=x{\vec{r}}_1+y{\vec{r}}_2$, where x and y are scalars. If the given vectors are linearly dependent then x and y will exist uniquely; otherwise not.;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Consider ;;;;;;${\vec{r}}_3=x{\vec{r}}_1+y{\vec{r}}_2$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ $(4\vec{a}-5\vec{b}+\vec{c})=x(2\vec{a}-3\vec{b}+\vec{c})+y(3\vec{a}-5\vec{b}+2\vec{c})$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ $4\vec{a}-5\vec{b}+\vec{c}=(2x+3y)\vec{a}+(-3x-5y)\vec{b}+(x+2y)\vec{c}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;but $\vec{a},\vec{b},\vec{c}$ are non–zero, non–coplanar vectors.;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Hence;;;;;;;;;;;2x+ 3y = 4;;;;;;;;;;;;;;;;;…(i);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;–3x – 5y = –5;;;;;;;;;;;…(ii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;x + 2y = 1;;;;;;;;;;;;;;;;;;…(iii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Solving (i) and (ii), we get x = 5, y = –2 which also satisfy (iii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;x and y are unique. $\Rightarrow$ ${\vec{r}}_3=5{\vec{r}}_1-2{\vec{r}}_2$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Hence;;;;;;;;;;;${\vec{r}}_1,{\vec{r}}_2$ and ${\vec{r}}_3$ are linearly dependent vectors.;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;ORTHOGONAL SYSTEM OF VECTORSIn the orthogonal system of vectors we choose these vectors as three mutually perpendicular unit vectors denoted by $\hat{i},\hat{j}$, and $\hat{k}$ directed along the positive directions of X, Y and Z axes respectively. Corresponding to any point P(x, y z) we can associate a vector w.r.t. a fixed orthogonal system and then this vector is the position vector (p.v.) of that point. i.e. p.v. of P = $x\hat{i}+y\hat{j}+z\hat{k}$
Distance of P from O = $|\overline{OP}|$= $\sqrt{x^2+y^2+z^2}.$ x, y, z are called the components of the vector If a vector $\overline{OP}$makes angles $\alpha$, $\beta$, $\gamma$ with the positive directions of X, Y and Z axes respectively, then cos$\alpha$, cos$\beta$, cos$\gamma$ are called the direction cosines (d.c.'s) of $\overline{OP}$.cos$\alpha$ = $\frac{x}{OP};$ cos$\beta$$\frac{y}{OP}$; cos$\gamma$ = $\frac{z}{OP}$

So that cos^2$\alpha$ + cos^2$\beta$ + cos^2$\gamma$ = 1

Unit vector in the direction of $\overline{OP}$is

$\frac{\overline{OP}}{OP}=\frac{x}{OP}\hat{i}+\frac{y}{OP}\hat{j}+\frac{z}{OP}\hat{k}$ = $\cos \alpha \hat{i}+\cos \beta \hat{j}+\cos \gamma \hat{k}$.

Section formula

INTERNAL DIVISION

Let A and B be two points with position vectors $\vec{a}$ and $\vec{b}$ respectively, and C be a point dividing AB internally in the ratio m : n. Then the position vector of C is given by $\overrightarrow{OC}=\frac{m\vec{b}+n\vec{a}}{m+n}$.

Proof: Let O be the origin. The $\overrightarrow{OA}=\vec{a}and\overrightarrow{OB}=\vec{b}$, let $\vec{c}$ be the position vector of C which divides AB internally in the ratio m : n then,

$\frac{AC}{CB}=\frac{m}{n}$

$\Rightarrow$ n.$\overrightarrow{AC}=m\cdot \overrightarrow{CB}$

$\Rightarrow$ n(P.V. of $\vec{C}$ – P.V. of $\vec{A}$) = m(P.V. of $\vec{B}$ – P.V. of $\vec{C}$)

$\Rightarrow$ $n(\vec{c}-\vec{a})=m(\vec{b}-\vec{c})$

$\Rightarrow$ $n\vec{c}-n\vec{a}=m\vec{b}-m\vec{c}$ $\Rightarrow$ $\vec{c}(n+m)=m\vec{b}+n\vec{a}$

$\Rightarrow$ $\vec{c}=\frac{m\vec{b}+n\vec{a}}{n+m}$ or $\overrightarrow{OC}=\frac{m\vec{b}+n\vec{a}}{m+n}$

EXTERNAL DIVISION

Let A and B be two points with position vectors $\vec{a}$ and $\vec{b}$ respectively and let C be a point dividing $\overrightarrow{AB}$ externally in the ratio m : n. Then the position vector of $\vec{C}$ is given by$\overrightarrow{OC}=\frac{m\vec{b}-n\vec{a}}{m-n}$.

Note:

(i) If C is the mid–point of AB, then P.V. of C is $\frac{\vec{a}+\vec{b}}{2}$.

(ii) We have,$\vec{c}=\frac{m\vec{b}+n\vec{a}}{m+n}=\frac{m}{m+n}\vec{b}+\frac{n}{m+n}\vec{a}$. Hence $\vec{c}$ is in the form of $\vec{c}=\lambda \vec{a}+\mu \vec{b}$.

where, $\lambda =\frac{n}{m+n}$ and $\mu =\frac{m}{m+n}$.Thus, position vector of any point $\vec{C}$ on $\overrightarrow{AB}$ can always be taken as $\vec{C}=\lambda \vec{a}+\mu \vec{b}$ where $\lambda$ + $\mu$ = 1.

(iii) If circumcentre is origin and vertices of a triangle have position vectors $\vec{a},\vec{b},\vec{c}$, then the position vector of orthocentre will be $-(\vec{a}+\vec{b}+\vec{c})$.

Illustration 3: ABC is a triangle. A line is drawn parallel to BC to meet AB and AC in D and E respectively. Prove that the median through A bisects DE.

Solution: Take the vertex A of the triangle ABC as the origin. Let $\bar{b}and\bar{c}$ be the p.v. of B and C. The mid point of BC has the p.v. = $\frac{\bar{b}+\bar{c}}{2}.$ The equation of the median is $r=t\frac{\bar{b}+\bar{c}}{2}$. Let D divide AB in the ratio 1:$\mu$

$\Rightarrow$ p.v. of $D=\frac{\bar{b}}{1+\mu }$..Let E divide AC in the ratio 1:$\lambda$

$\Rightarrow$ p.v. E = $\frac{\bar{c}}{1+\lambda }$ $\Rightarrow$ $\lambda$ = $\mu$sz p.v. of the mid-point of DE = $\frac{\bar{b}+\bar{c}}{2(1+\mu )}$which lies on the median. Hence the median bisects DE.

Bisector of the angle between two vectors

Consider two non–zero, non–collinear vectors $\bar{a}$ and $\bar{b}$. The bisector of the angle between the two vectors $\bar{a}$ and $\bar{b}$ is k$\left(\frac{\bar{a}}{|\bar{a}|}+\frac{\bar{b}}{|\bar{b}|}\right)$

where k $\in$ R⁺.

Illustration 4: If the vector $\hat{i}-3\hat{j}+5\hat{k}$ bisects the angle between $\hat{a}$ and $-\hat{i}+2\hat{j}+2\hat{k}$, where $\hat{a}$ is a unit vector then find $\hat{a}$ .

Solution: According to the given conditions $\lambda$$(\hat{i}-3\hat{j}+5\hat{k})$ = $\hat{a}+\frac{2\hat{k}+2\hat{j}-\hat{i}}{3}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;3$\hat{a}$ = 3$\lambda$$(\hat{i}-3\hat{j}+5\hat{k})-(2\hat{k}+2\hat{j}-\hat{i})$

= $\hat{i}$(3$\lambda$ + 1) – $\hat{j}$(2 + 9$\lambda$) + $\hat{k}$(15$\lambda$ – 2)

$\Rightarrow$ $|\hat{a}|=\sqrt{{(3\lambda +1)}^2+{(2+9\lambda )}^2+{(15\lambda -2)}^2}$

$\Rightarrow$ 9 = (3$\lambda$ + 1)² + (2 + 9$\lambda$)² + (15$\lambda$ – 2)²

$\Rightarrow$ 315$\lambda$² – 18$\lambda$ = 0 $\Rightarrow$ $\lambda$ = 0, $\frac{2}{35}$.

If $\lambda$ = 0, $\hat{a}=\hat{i}-2\hat{j}-2\hat{k}$ (not acceptable)

For $\lambda$ = $\frac{2}{35}$, $\hat{a}=\frac{1}{105}(41\hat{i}-88\hat{j}-40\hat{k})$