# 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

$\overrightarrow{AB}$
$\color {purple} {Example\ 1\ .}\ \overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}\ and\ \overrightarrow{b}\ =\ \overrightarrow{j}\ – 2\ \overrightarrow{k},\ \color {red} {find\ \overrightarrow{a}\ -\ 2\overrightarrow{b}}\ \hspace{15cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{b}\ =\ \overrightarrow{j}\ – 2\ \overrightarrow{k}$
$\overrightarrow{a}\ -\ 2\overrightarrow{b}\ =\ 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\ -\ 2(\overrightarrow{j}\ -\ 2\overrightarrow{k})$
$=\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{a}\ -\ 2\overrightarrow{b}\ =\ 3\overrightarrow{i}\ +\ 5\overrightarrow{k}\ \hspace{5cm}$

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.

$\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}$
$OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }= r$
$cos\ α\ =\ \frac{x}{r},\ cos\ β\ =\ \frac{y}{r},\ cos\ γ\ =\ \frac{z}{r}$
$Direction\ cosines\ are \frac{x}{r}, \frac{y}{r}, \frac{z}{r}$
$Result:\ for\ any\ vector\ the\ sum\ of\ squares\ of\ direction\ cosines\ of\ \overrightarrow{r}\ is\ 1$
$\hspace{2cm}\ i.e\ cos^2\ α\ +\ cos^2\ β\ +\ cos^2\ γ\ =\ 1$
$Direction\ ratio\ are\ x,\ y,\ z$
$\color {purple} {Example\ 2\ .}\ \color {red} {Find\ the\ Direction\ cosines\ and\ Direction\ ratios}\ of\ the\ vector\ 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\overrightarrow{a}= 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}$
$r =\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2 + (-5)^2 }$
$= \sqrt{(9 + 16 +25 }$
$r =\sqrt{50}$
$Direction\ cosines\ are \frac{3}{\sqrt(50)}, \frac{4}{\sqrt(50)}, \frac{-5}{\sqrt(50)}$
$Direction\ ratios\ are\ 3, 4, -5$

$\color {purple} {Example\ 3\ .}\ \color {red} {Can\ a\ vector\ have\ direction\ angles}\ 30^0,\ 45^0,\ 60^0\ \hspace{10cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$The\ condition\ is\ cos^2\ α\ +\ cos^2\ β\ +\ cos^2\ γ\ =\ 1$
$Here\ α\ =\ 30^0,\ β\ =\ 45^0,\ γ\ =\ 60^0$
$cos^2\ α\ +\ cos^2\ β\ +\ cos^2\ γ\ =\ cos^2\ 30^0\ +\ cos^2\ 45^0\ +\ cos^2\ 60^0$
$=\ (\frac{\sqrt{3}}{2})^2\ +\ (\frac{1}{\sqrt{2}})^2\ +\ (\frac{1}{2})^2$
$=\ \frac{3}{4}\ +\ \frac{1}{2}\ +\ \frac{1}{4}$
$=\ \frac{3\ +\ 2\ +\ 1}{4}$
$=\ \frac{6}{4}\ \neq 1$
$\boxed {\therefore\ they\ are\ not\ directions\ angles\ of\ any\ vector}$

#### 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

$\overrightarrow{OA}= x_1\overrightarrow{i}\ + y_1\overrightarrow{j}+ z_1\overrightarrow{k}$
$\overrightarrow{OB}= x_2\overrightarrow{i}\ + y_2\overrightarrow{j}+ z_2\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \overrightarrow{|OB – BA|} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 +(z_2 – z_1)^2 }$
$\color {purple} {Example\ 4\ .}\ If\ position\ vectors\ of\ the\ points\ A\ and\ B\ are\ \hspace{8cm}$$\overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k}\ and\ 2\overrightarrow{i}\ – \overrightarrow{j},\ \color {red} {find\ \overrightarrow{|AB|}}\ \hspace{6cm}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}= \overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}\ – \overrightarrow{j}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=2\overrightarrow{i}\ – \overrightarrow{j}- (\overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k})$
$=2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{i}\ – 2\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} – 3\overrightarrow{j} + 3\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(1)^2 + (-3)^2 +(3)^2 }$
$= \sqrt{(1 + 9 +9 })$
$= \sqrt{19}$

Defn: If a is a vector,  then

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

$\color {purple} {Example\ 5\ .}\ \color {red} {Find\ the\ Unit\ vector\ along\ the\ vector}\ 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}\ \hspace{15cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\overrightarrow{a}= 3\overrightarrow{i} + 4\overrightarrow{j} – 5\overrightarrow{k}$
$\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2+(-5)^2 }$
$= \sqrt{(9 + 16 +25}$
$=\sqrt{50}$
$\overrightarrow{|a|}=\sqrt{50}$
$Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}}{\sqrt{50}}$

$\underline{\color {red} {Condition\ for\ two\ position\ vectors\ \overrightarrow{a}\ and\ \overrightarrow{b}\ to\ be\ collinear}}\ if\ \color {green} {\overrightarrow{a}\ =\ k\ \overrightarrow{b}}\ \hspace{10cm}$
$\underline{\color {red} {Condition\ for\ three\ position\ vectors\ \overrightarrow{OA}\ ,\ \overrightarrow{OB}\ and\ \overrightarrow{OC}\ to\ be}}\ \hspace{10cm}$
$i)\ \color {green} {collinear \ if \overrightarrow{BC} = K\overrightarrow{AB}}$
$ii) \color {green} {Equilateral\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}=\overrightarrow{|AC|}}$
$iii) \color {green} {Isosceles\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}\not=\overrightarrow{|AC|} or\ \overrightarrow{|BC|}=\overrightarrow{|AC|}\not=\overrightarrow{|AB|}}$
$iv) \color {green} {right\ angled\ triangle\ if\ (AB)^2+(BC)^2=(AC)^2 or\ (BC)^2+(AC)^2=(AB)^2}$
$\color {purple} {Example\ 6\ .}\ If\ the\ vectors\ 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ and\ 6\overrightarrow{i}\ -\ m\overrightarrow{j}\ are\ collinear,\ \hspace{10cm}$$\color {red} {find\ the\ value\ of\ m}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{19cm}$
$Given\ \hspace{15cm}$
$\overrightarrow{a}\ =\ 2\overrightarrow{i}\ – 3\overrightarrow{j}\ and\ \overrightarrow{b}\ =\ 6\overrightarrow{i}\ -\ m\overrightarrow{j}$
$Given\ that\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ collinear$
$\overrightarrow{a}\ =\ k\ \overrightarrow{b}$
$2\overrightarrow{i}\ – 3\overrightarrow{j}\ =\ k\ (6\overrightarrow{i}\ -\ m\overrightarrow{j})$
$Comparing\ LHS\ and\ RHS\ we\ get$
$2\ =\ 6k\ — (1)\ \hspace{2cm}\ and\ \hspace{2cm}\ -\ 3\ =\ -\ mk\ — (2)$
$k\ =\ \frac{1}{3}$
$Put\ k\ value\ in\ (2)\ we\ get$
$-\ 3\ =\ -\ m (\frac{1}{3})$
$m\ =\ 9$

$\color {purple} {Example\ 7\ .}\ \color {red} {Show\ that\ the\ points\ whose\ position\ vectors}\ \hspace{12cm}$$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}\ \color {red} {are\ collinear}\ \hspace{5cm}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}= 2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}\ +\overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{OC}= 6\overrightarrow{i}\ – 5\overrightarrow{j} + 7\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}+ \overrightarrow{j} – 2\overrightarrow{k}- (2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k})$
$=3\overrightarrow{i}+ \overrightarrow{j} – 2\overrightarrow{k}- 2\overrightarrow{i}\ – 3\overrightarrow{j}+ 5\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=6\overrightarrow{i}- 5 \overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}\ + \overrightarrow{j}- 2\overrightarrow{k})$
$=6\overrightarrow{i}-5 \overrightarrow{j} +7\overrightarrow{k}- 3\overrightarrow{i}\ – \overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{BC}= 3\overrightarrow{i} – 6\overrightarrow{j} + 9\overrightarrow{k}$
$=\ 3( \overrightarrow{i}\ – 2\overrightarrow{j}\ +\ 3\overrightarrow{k})$
$\overrightarrow{BC}\ =\ 3\ \overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$

$\color {purple} {Example\ 8\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}$$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}\ \color {red} {form\ an\ equilateral\ triangle}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}= 4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}\ +3\overrightarrow{j} + 4\overrightarrow{k}$
$\overrightarrow{OC}= 3\overrightarrow{i}\ + 4\overrightarrow{j} + 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})$
$=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{AB}= -2\overrightarrow{i} +\overrightarrow{j} + \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(-2)^2 + (-1)^2 +(1)^2 }$
$= \sqrt{(4 + 1 +1 }$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (2\overrightarrow{i}+ 3\overrightarrow{j}+ 4\overrightarrow{k})$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 2\overrightarrow{i}- 3\overrightarrow{j}- 4\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }$
$= \sqrt{(1 + 1 + 4}$
$BC = \sqrt{6}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})$
$=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{AC}= -\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(-1)^2 + (2)^2 +(-1)^2 }$
$= \sqrt{(1 + 4 + 1}$
$AC = \sqrt{6}$
$AB = BC = AC = \sqrt{6}$

The given triangle is an equilateral triangle.

$\color {purple} {Example\ 9\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}$$2\overrightarrow{j}\ +\ 10 \overrightarrow{k}\ ,\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k} and\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ \color {red} {form\ an\ isosceles\ triangle}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}\ =\ 2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k}$
$\overrightarrow{OB}\ =\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}$
$\overrightarrow{OC}\ =\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}\ – (2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k})$
$=\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ 2 \overrightarrow{j}\ -\ 10\overrightarrow{k}$
$\overrightarrow{AB}\ =\ 7\overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ 4\ \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(7)^2\ +\ (4)^2\ +\ (-4)^2 }$
$= \sqrt{(49\ +\ 16\ +\ 16)}$
$=\ \sqrt{81}$
$AB =\ 9$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ – (7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k})$
$=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ -\ 7\overrightarrow{i}\ – 6\ \overrightarrow{j}\ -\ 6\overrightarrow{k}$
$\overrightarrow{BC}\ =\ -\ 11\overrightarrow{i}\ +\ 3\overrightarrow{j}$
$BC =\overrightarrow{|BC|} = \sqrt{(-11)^2 + (3)^2}$
$= \sqrt{121\ +\ 9}$
$=\ \sqrt{130}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ – (2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k})$
$=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ -\ 2\ \overrightarrow{j}\ -\ 10\overrightarrow{k}$
$\overrightarrow{AC}\ =\ -\ 4\overrightarrow{i}\ +\ 7\overrightarrow{j}\ -\ 4\ \overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(-\ 4)^2\ +\ (7)^2\ +\ (-\ 4)^2 }$
$= \sqrt{16\ +\ 49\ +\ 16}$
$AC = \sqrt{81}$
$\boxed{AB = AC\ \neq\ BC}$
$The\ given\ triangle\ is\ isosceles\ triangle$

$\color {purple} {Example\ 10\ .}\ \color {red} {Prove\ that\ the\ points\ whose\ position\ vectors\ are}\ \hspace{15cm}$$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}\ \color {red} {form\ a\ right\ angled\ triangle}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}= 3\overrightarrow{i} – \overrightarrow{j}+ 6\overrightarrow{k}$
$\overrightarrow{OB}= 5\overrightarrow{i}\ -2\overrightarrow{j} + 7\overrightarrow{k}$
$\overrightarrow{OC}= 6\overrightarrow{i}\ – 5\overrightarrow{j} + 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})$
$=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- 3\overrightarrow{i}+ \overrightarrow{j}- 6\overrightarrow{k}$
$\overrightarrow{AB}= 2\overrightarrow{i} -\overrightarrow{j} + \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(2)^2 + (-1)^2 +(1)^2 }$
$= \sqrt{(4 + 1 +1 }$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (5\overrightarrow{i}- 2\overrightarrow{j}+ 7\overrightarrow{k})$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 5\overrightarrow{i}+ 2\overrightarrow{j}- 7\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} -3\overrightarrow{j} -5\overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-3)^2 +(-5)^2 }$
$= \sqrt{(1 + 9 + 25}$
$BC = \sqrt{35}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})$
$=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 3\overrightarrow{i}+\overrightarrow{j}- 6\overrightarrow{k}$
$\overrightarrow{AC}= 3\overrightarrow{i} -4\overrightarrow{j} -4\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(3)^2 + (-4)^2 +(-4)^2 }$
$= \sqrt{(9 + 16 + 16}$
$AC = \sqrt{41}$
$AB^2 = 6, BC^2 = 35, AC^2 = 41$
$AB^2 + BC^2 = 6 + 35 = 41=AC^2$
$AB^2 + BC^2 = AC^2$

The given triangle is an right angled triangle.

$Let\ \overrightarrow{a}\ =\ a_1\ \overrightarrow{i}\ +\ a_2\ \overrightarrow{j}\ +\ a_3\ \overrightarrow{k}\ \hspace{10cm}$
$\overrightarrow{b}\ =\ b_1\ \overrightarrow{i}\ +\ b_2\ \overrightarrow{j}\ +\ b_3\ \overrightarrow{k}\ \hspace{8cm}$
$\overrightarrow{c}\ =\ c_1\ \overrightarrow{i}\ +\ c_2\ \overrightarrow{j}\ +\ c_3\ \overrightarrow{k}\ \hspace{8cm}$
$Then\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ =\begin{vmatrix} a_1 &amp: a_2 &amp: a_3 \\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$
$\underline{\color {red} {Condition\ for\ three\ vectors\ \overrightarrow{a}\ \overrightarrow{b}\ and\ \overrightarrow{c}\ to\ be\ coplanar}}\ if\ \color {green} {[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ =\ 0}\ \hspace{10cm}$
$\underline{\color {red} {Condition\ for\ three\ position\ vectors\ \overrightarrow{OA}\ ,\ \overrightarrow{OB}\ \overrightarrow{OC}\ and\ \overrightarrow{OD}\ to\ be\ coplanar}}\ if\ \color {green} {[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\ =\ 0}\ \hspace{10cm}$
$\color {purple} {Example\ 11\ .}\ \color {red} {Find\ the\ value\ of\ p}\ such\ that\ the\ vectors\ \hspace{15cm}$$2\overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\overrightarrow{k}, p\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k} and\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}\ \color {red} {lie\ on\ the\ same\ plane}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{a}\ =\ 2\overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\overrightarrow{k}$
$\overrightarrow{b}\ =\ p\overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ 6\overrightarrow{k}$
$\overrightarrow{c}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}$
$Given\ \overrightarrow{a}\ \overrightarrow{b}\ and\ \overrightarrow{c}\ lie\ on\ the\ same\ plane$
$[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ =\ 0$
$\begin{vmatrix} 2 & -3 & 5 \\ p & 2 & -6 \\ 3 & -1 & 4 \\ \end{vmatrix}\ =\ 0$
$2\begin{vmatrix} 2 & -6 \\ -1 & 4 \\ \end{vmatrix}\ +\ 3\begin{vmatrix} p & -6 \\ 3 & 4\\ \end{vmatrix}\ +\ 5\begin{vmatrix} p & 2 \\ 3 & – 1\\ \end{vmatrix}\ =\ 0$
$2(8\ -\ 6)\ +\ 3 (4p\ +\ 18)\ +\ 5(-p\ -\ 6)\ =\ 0$
$4\ +\ 12\ p\ +\ 54\ -\ 5\ p\ -\ 30\ =\ 0$
$7p\ +\ 58\ -\ 30\ =\ 0$
$7p\ +\ 28\ =\ 0$
$7p\ =\ -\ 28$
$\boxed{p\ =\ -\ 4}$

$\color {purple} {Example\ 12\ .}\ \color {red} {Prove\ that\ the\ points\ given\ by\ the\ position\ vectors}\ \hspace{15cm}$$4\overrightarrow{i}\ +\ 5\ \overrightarrow{j}\ +\ \overrightarrow{k}, -\ \overrightarrow{j}\ -\ \overrightarrow{k}, 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ and\ -4\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \color {red} {are\ coplanar}$
$\color {blue} {Soln\ :}\ Given\ \hspace{17cm}$
$\overrightarrow{OA}\ =\ 4\overrightarrow{i}\ +\ 5\ \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{OB}\ =\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OC}\ =\ 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{OD}\ =\ -4\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (4\overrightarrow{i}\ +\ 5\ \overrightarrow{j}\ +\ \overrightarrow{k})$
$=\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 4\overrightarrow{i}\ -\ 5\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{AB}\ =\ -\ 4\overrightarrow{i}\ -\ 6\overrightarrow{j}\ -\ 2\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ -\ (-\ \overrightarrow{j}\ -\ \overrightarrow{k})$
$=\ 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ +\ \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{BC}\ =\ 3\ \overrightarrow{i}\ +\ \ 10\overrightarrow{j}\ +\ 5\overrightarrow{k}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=\ 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ -\ (4\overrightarrow{i}\ +\ 5\ \overrightarrow{j}\ +\ \overrightarrow{k})$
$=\ 3\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ -\ 4\overrightarrow{i}\ -\ 5\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{AC}\ =\ -\ \overrightarrow{i}\ +\ 4\ \overrightarrow{j}\ +\ 3\ \overrightarrow{k}$
$\overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA}$
$=\ -4\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 4\overrightarrow{k}\ -\ (4\overrightarrow{i}\ +\ 5\ \overrightarrow{j}\ +\ \overrightarrow{k})$
$=\ -4\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 4\overrightarrow{k}\ -\ 4\overrightarrow{i}\ -\ 5\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{AD}\ =\ -\ 8\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 3\ \overrightarrow{k}$
$To\ claim\ \overrightarrow{AB}\ \overrightarrow{AC}\ and\ \overrightarrow{AD}\ coplanar\ need\ to\ prove\ [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\ =\ 0$
$[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]\ =\ \begin{vmatrix} – 4 & – 6 & – 2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \\ \end{vmatrix}$
$=\ -4\begin{vmatrix} 4 & 3 \\ -1 & 3 \\ \end{vmatrix}\ +\ 6\begin{vmatrix} -1 & 3 \\ -8 & 3\\ \end{vmatrix}\ -\ 2\begin{vmatrix} -1 & 4 \\ -8 & – 1\\ \end{vmatrix}$
$=\ -4\ (12\ +\ 3)\ +\ 6 (-3\ +\ 24)\ -\ 2(1\ +\ 32)$
$=\ -\ 4\ (15)\ +\ 6\ (21)\ -\ 2\ (33)$
$=\ -\ 60\ +\ 126\ -\ 66$
$=\ -\ 126\ +\ 126$
$=\ 0$
$\therefore\ the\ given\ position\ vectors\ \overrightarrow{AB}\ \overrightarrow{AC}\ and\ \overrightarrow{AD}\ are\ coplanar$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \overrightarrow{a}= 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k},\ \color {red} {find\ 3\overrightarrow{a}\ +\ \overrightarrow{b}}\ \hspace{10cm}$
$\color {purple} {2\ .}\ \overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k},\ \color {red} {find\ 2\overrightarrow{a}\ +\ 3\overrightarrow{b}}\ \hspace{10cm}$
$\color {purple} {3\ .}\ If\ position\ vectors\ of\ the\ points\ A\ and\ B\ are\ 2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k}\ and\ 5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k},\ \color {red} {find\ \overrightarrow{|AB|}}\ \hspace{20cm}$
$\color {purple} {4\ .}\ \color {red} {Find\ the\ unit\ vector\ along\ the\ vector}\ 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}\ \hspace{20cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {5\ .}\ If\ the\ position\ vector\ of\ the\ points\ A\ and\ B\ are\ \hspace{15cm}$$\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k},\ \hspace{12cm}$$\color {red} {find\ \overrightarrow{|AB|}\ ,\ Also\ find\ the\ direction\ ratio\ of\ \overrightarrow{AB}}\ \hspace{10cm}$
$\color {purple} {6\ .}\ \color {red} {Find\ the\ Modulus\ and\ Direction\ cosines}\ \hspace{15cm}$$of\ the\ vector\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{12cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {7\ .}\ \color {red} {\ Show\ that}\ the\ points\ whose\ position\ vectors\ \hspace{15cm}$$\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ \overrightarrow{k},\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}\ and\ -\ 7 \overrightarrow{j}\ -\ 5\overrightarrow{k}\ \color {red} {are\ collinear}\ \hspace{5cm}$
$\color {purple} {8\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}$$2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k}, 3\overrightarrow{i}\ + 4\overrightarrow{j}\ + 2\overrightarrow{k} and\ 4\overrightarrow{i}\ +\ 2 \overrightarrow{j}+ 3\overrightarrow{k}\ \color {red} {form\ an\ equilateral\ triangle}$
$\color {purple} {9\ .}\ \color {red} { Prove\ that\ the\ points}\ \hspace{15cm}$$3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2 \overrightarrow{k}\ ,\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k} and\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ \color {red} {form\ an\ isosceles\ triangle}$