# 2.1 VECTOR – INTRODUCTION

Vectors constitute one of the several Mathematical systems which can be usefully employed to provide mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied Mathematics. Vectors facilitate mathematical study of such physical quantities as possess Direction in addition to Magnitude. Velocity of a particle, for example, is one such quantity.

Physical quantities are broadly divided in two categories viz (a) Vector Quantities & (b) Scalar quantities.

( a ) Vector quantities :

Any quantity, such as velocity, momentum, or force, that has both magnitude and direction

( b ) Scalar quantities :

A quantity, such as mass, length, time, density or energy, that has size or magnitude but does not involve the concept of direction is called scalar quantity.

Mathematical Description of Vector: A vector is a directed line segment. The length of the segment is called magnitude of the vector. The direction is indicated by an arrow joining the initial and final points of the line segment. The vector AB i.e, joining the initial point A and the final point B in the direction of AB is denoted as

$\overrightarrow{AB}$

Triangle Law of  Addition of Two vectors:

$\overrightarrow{OA} = \overrightarrow{a}$ $\overrightarrow{AB} = \overrightarrow{b}$$\overrightarrow{OA}+ \overrightarrow{AB} = \overrightarrow{OB}$$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$

Position Vector: If P is any point in the space and 0 is the origin, then

$\overrightarrow{OP}$

is called the position vector of the point  P.

Let P be a point in a Three dimensional Space. Let 0 be the origin and i, j and k  the unit vectors along the x ,yand  z  axes . Then if P is (x, y, z) , the position vector of the point P is

$\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}$
$OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }$

#### Direction Cosinesand Direction ratios

When a directed line OP passing through the origin makes α, β and γ angles with the x, y and z axis respectively with O as the reference, these angles are referred as the direction angles of the line and the cosine of these angles give us the direction cosines. These direction cosines are usually represented as l, m and n.

$\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}$
$OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }= r$
$Direction\ cosines\ are \frac{x}{r}, \frac{y}{r}, \frac{z}{r}$

Direction ratio are  x, y, z

Example:

$Find\ the\ Direction\ cosines\ and\ Direction\ ratios\ of\ the\ vector\ 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}$

Soln:

$\overrightarrow{a}= 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}$
$r =\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2 + (-5)^2 }$
$= \sqrt{(9 + 16 +25 }$
$r =\sqrt{50}$
$Direction\ cosines\ are \frac{3}{\sqrt(50)}, \frac{4}{\sqrt(50)}, \frac{-5}{\sqrt(50)}$
$Direction\ ratios\ are 3, 4, -5$

#### Distance between two points:

If A and B are two points in the space with co-ordinates A (x1, y1, z1 ) and B (x2, y2, z2), then the position vectors are

$\overrightarrow{OA}= x_1\overrightarrow{i}\ + y_1\overrightarrow{j}+ z_1\overrightarrow{k}$
$\overrightarrow{OB}= x_2\overrightarrow{i}\ + y_2\overrightarrow{j}+ z_2\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \overrightarrow{|OB - BA|} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 +(z_2 - z_1)^2 }$

Example  :   If  position vectors of the points A and B are

$3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}and\overrightarrow{i}\ - \overrightarrow{j}+ 3\overrightarrow{k}$
$find \overrightarrow{|AB|}$

Sol.:     Given

$\overrightarrow{OA}= 3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{OB}= \overrightarrow{i}\ - \overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\overrightarrow{i}\ - \overrightarrow{j} + 3\overrightarrow{k}- (3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k})$
$=\overrightarrow{i}\ - \overrightarrow{j} + 3\overrightarrow{k}- 3\overrightarrow{i}\ - 2\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{AB}= - 2\overrightarrow{i} - 3\overrightarrow{j} + 4\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(-2)^2 + (-3)^2 +(4)^2 }$
$= \sqrt{(4 + 9 +16 }$
$= \sqrt{29}$

If a is a vector,  then

$Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}$

Example  :   Find the unit vector along the vector

$3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}$

Soln:

$\overrightarrow{a}= 3\overrightarrow{i} + 4\overrightarrow{j} - 5\overrightarrow{k}$
$\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2+(-5)^2 }$
$= \sqrt{(9 + 16 +25}$
$=\sqrt{50}$
$\overrightarrow{|a|}=\sqrt{50}$
$Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}}{\sqrt{50}}$

Condition for three position vectors

$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ be\ collinear \ if \overrightarrow{BC} = K\overrightarrow{AB}$

Example  : Show that the points whose position vectors

$2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k}, 3\overrightarrow{i}\ + \overrightarrow{j}- 2\overrightarrow{k} and\ 6\overrightarrow{i}\ -5 \overrightarrow{j}+ 7\overrightarrow{k}\ are\ collinear$

Soln:   Given

$\overrightarrow{OA}= 2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}\ +\overrightarrow{j} - 2\overrightarrow{k}$
$\overrightarrow{OC}= 6\overrightarrow{i}\ - 5\overrightarrow{j} + 7\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}+ \overrightarrow{j} - 2\overrightarrow{k}- (2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k})$
$=3\overrightarrow{i}+ \overrightarrow{j} - 2\overrightarrow{k}- 2\overrightarrow{i}\ - 3\overrightarrow{j}+ 5\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=6\overrightarrow{i}- 5 \overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}\ + \overrightarrow{j}- 2\overrightarrow{k})$
$=6\overrightarrow{i}-5 \overrightarrow{j} +7\overrightarrow{k}- 3\overrightarrow{i}\ - \overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{BC}= 3\overrightarrow{i} - 6\overrightarrow{j} + 9\overrightarrow{k}$
$\overrightarrow{BC}= 3(\overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k})$
$\overrightarrow{BC} = 3\overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$

Home work problem

Show that the points whose position vectors

$2\overrightarrow{i}\ - \overrightarrow{j}+ 3\overrightarrow{k}, 3\overrightarrow{i}\ - 5\overrightarrow{j}+ \overrightarrow{k} and\ -\overrightarrow{i}\ + 11 \overrightarrow{j}+ 9\overrightarrow{k}\ are\ collinear$

Soln:   Given

$\overrightarrow{OA}= 2\overrightarrow{i} - \overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i} - 5\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{OC}= -\overrightarrow{i} + 11\overrightarrow{j} + 9\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}- 5\overrightarrow{j} + \overrightarrow{k}- (2\overrightarrow{i}\ - \overrightarrow{j}+ 3\overrightarrow{k})$
$=3\overrightarrow{i}- 5\overrightarrow{j} + \overrightarrow{k}- 2\overrightarrow{i}\ + \overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} - 4\overrightarrow{j} - 2\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=-\overrightarrow{i}+11 \overrightarrow{j} + 9\overrightarrow{k}- (3\overrightarrow{i}\ - 5\overrightarrow{j}+ \overrightarrow{k})$
$=-\overrightarrow{i}+11 \overrightarrow{j} + 9\overrightarrow{k}- 3\overrightarrow{i}\ + 5\overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{BC}= -4\overrightarrow{i} + 16\overrightarrow{j} + 8\overrightarrow{k}$
$\overrightarrow{BC}= -4(\overrightarrow{i} - 4\overrightarrow{j} - 2\overrightarrow{k})$
$\overrightarrow{BC} = -4\overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$
$Condition for\ three\ position\ vectors\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ be$
$i) Equilateral\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}=\overrightarrow{|AC|}$
$ii) Isosceles\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}\not=\overrightarrow{|AC|} or\ \overrightarrow{|BC|}=\overrightarrow{|AC|}\not=\overrightarrow{|AB|}$
$iii) right\ angled\ triangle\ if\ (AB)^2+(BC)^2=(AC)^2 or\ (BC)^2+(AC)^2=(AB)^2$
1. Prove that the points
$4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}, 2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k} and\ 3\overrightarrow{i}\ +4 \overrightarrow{j}+ 2\overrightarrow{k}\ form\ an\ equilateral\ triangle$

Soln: Given

$\overrightarrow{OA}= 4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}\ +3\overrightarrow{j} + 4\overrightarrow{k}$
$\overrightarrow{OC}= 3\overrightarrow{i}\ + 4\overrightarrow{j} + 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})$
$=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{AB}= -2\overrightarrow{i} +\overrightarrow{j} + \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(-2)^2 + (-1)^2 +(1)^2 }$
$= \sqrt{(4 + 1 +1 }$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (2\overrightarrow{i}+ 3\overrightarrow{j}+ 4\overrightarrow{k})$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 2\overrightarrow{i}- 3\overrightarrow{j}- 4\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }$
$= \sqrt{(1 + 1 + 4}$
$BC = \sqrt{6}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{AC}= -\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(-1)^2 + (2)^2 +(-1)^2 }$
$= \sqrt{(1 + 4 + 1}$
$AC = \sqrt{6}$
$AB = BC = AC = \sqrt{6}$

The given triangle is an equilateral triangle.

2. Prove that the points whose position  vectors  are

$3\overrightarrow{i}\ - \overrightarrow{j}+ 6\overrightarrow{k}, 5\overrightarrow{i}\ - 2\overrightarrow{j}+ 7\overrightarrow{k} and\ 6\overrightarrow{i}\ -5 \overrightarrow{j}+ 2\overrightarrow{k}\ form\ a\ right\ angled\ triangle$

Soln: Given

$\overrightarrow{OA}= 3\overrightarrow{i} - \overrightarrow{j}+ 6\overrightarrow{k}$
$\overrightarrow{OB}= 5\overrightarrow{i}\ -2\overrightarrow{j} + 7\overrightarrow{k}$
$\overrightarrow{OC}= 6\overrightarrow{i}\ - 5\overrightarrow{j} + 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})$
$=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- 3\overrightarrow{i}+ \overrightarrow{j}- 6\overrightarrow{k}$
$\overrightarrow{AB}= 2\overrightarrow{i} -\overrightarrow{j} + \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(2)^2 + (-1)^2 +(1)^2 }$
$= \sqrt{(4 + 1 +1 }$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (5\overrightarrow{i}- 2\overrightarrow{j}+ 7\overrightarrow{k})$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 5\overrightarrow{i}+ 2\overrightarrow{j}- 7\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} -3\overrightarrow{j} -5\overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-3)^2 +(-5)^2 }$
$= \sqrt{(1 + 9 + 25}$
$BC = \sqrt{35}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 3\overrightarrow{i}+\overrightarrow{j}- 6\overrightarrow{k}$
$\overrightarrow{AC}= 3\overrightarrow{i} -4\overrightarrow{j} -4\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(3)^2 + (-4)^2 +(-4)^2 }$
$= \sqrt{(9 + 16 + 16}$
$AC = \sqrt{41}$
$AB^2 = 6, BC^2 = 35, AC^2 = 41$
$AB^2 + BC^2 = 6 + 35 = 41=AC^2$
$AB^2 + BC^2 = AC^2$

The given triangle is an right angled triangle.