# 2.2 PRODUCT OF VECTORS

#### SCALAR PRODUCT

##### Definition:
$Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.$ $Then\ the\ scalar\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by$$\overrightarrow{a}.\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\cos\theta$
##### Properties of Scalar Product:
$1. \overrightarrow{a}\ and\overrightarrow{b}are\ perpendicular\ vectors\ if\ and\ only\ if \overrightarrow{a}.\overrightarrow{b}=0.$
$2. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively$
$Then\ i) \overrightarrow{i}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1$
$ii) \overrightarrow{i}.\overrightarrow{j}= \overrightarrow{j}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{i}= \overrightarrow{k}.\overrightarrow{j}=\overrightarrow{i}.\overrightarrow{k}=0$
$3. \ Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}$
$4. \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$If\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then$
$\overrightarrow{a}.\overrightarrow{b}= a_1b_1 + a_2b_2 + a_3b_3\ (using\ property\ 2)$
$\color {purple} {Example\ 1\ .}\ \color {red} {Find\ the\ scalar\ product\ of}\ 2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}\ and\ \overrightarrow{i}\ +6\overrightarrow{j}+ 12\overrightarrow{k}\ \hspace{10cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\overrightarrow{a}\ =\ 2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}\ +\ 6\overrightarrow{j}+ 12\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}) .( \overrightarrow{i} +6\overrightarrow{j}+ 12\overrightarrow{k})$

= 2(1) – 4 (6) + 8 (12)

= 2 – 24 + 96

= 74

$\boxed{\overrightarrow{a}.\overrightarrow{b}= 74}$

$\color {purple} {Example\ 2\ .}\ \color {red} {Find\ the\ Dot\ product\ of}\ \overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}\ ,\ \overrightarrow{i}\ + 3\overrightarrow{k}\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= \overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}) .(\overrightarrow{i}\ + 3\overrightarrow{k})$
$=\ 1(1)\ +\ 1(0)\ +\ 1(3)$
$=\ 1\ +\ 0\ +\ 3$
$=\ 4$
$\boxed{\overrightarrow{a}.\overrightarrow{b}\ =\ 4}$

$\color {purple} {Example\ 3\ .}\ \color {red} {Write\ the\ condition\ for\ two\ vectors\ to\ be\ perpendicular}\ \hspace{15cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}.\overrightarrow{b}\ =\ 0$
$\color {purple} {Example\ 4\ .}\ \color {red} {Show\ that\ the\ vectors}\ 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k} and\ 3\overrightarrow{i}+ 2\overrightarrow{j}+6\overrightarrow{k} are\ perpendicular\ to\ each\ other$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{b}= 3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k}) .(3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k})$

= 2(3) + 3(2) – 2 (6)

= 6 + 6 -12

= 0

$\therefore\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.$

$\color {purple} {Example\ 5\ .}\ \color {red} {Find\ the\ value\ of\ m}\ if\ the\ vectors\ 3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k} and\ -6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k} are\ perpendicular\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k}$
$\overrightarrow{b}= -6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k}$
$Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other$
$i.e\ \overrightarrow{a}.\overrightarrow{b}= 0$
$(3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k}).(-6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k}) = 0 .$

3(-6) – 1(m) + 5 (4) = 0

-18 – m + 20 = 0

-m + 2 = 0

$\boxed{m\ =\ 2}$

$\color {purple} {Example\ 6\ .}\ \color {red} {\ Prove\ that}\ the\ vectors\ \overrightarrow{i}-\overrightarrow{j}+ 2\overrightarrow{k},\ \overrightarrow4{j}+ 2\overrightarrow{k}and\ -10\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k} are\ mutually\ perpendicular.\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= \overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{b}= + 4\overrightarrow{j}+2\overrightarrow{k}$
$\overrightarrow{c}= -10\overrightarrow{i}- 2\overrightarrow{j}+ 4\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k}) .(4\overrightarrow{j}+ 2\overrightarrow{k})$

= 1(0) – 1(4) + 2 (2)

= 0

$\boxed{\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors}$
$\overrightarrow{b}.\overrightarrow{c}= (0\overrightarrow{i}+ 4\overrightarrow{j}+ 2\overrightarrow{k}) .(-10\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})$

= 0(-10) + 4(-2) + 2

= 0 – 8 + 8

= 0

$\boxed{\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors}$
$\overrightarrow{c}.\overrightarrow{a}= (-10\overrightarrow{i}- 2\overrightarrow{j}+ 4\overrightarrow{k}) .(\overrightarrow{i}-\overrightarrow{j}+ 2\overrightarrow{k})$

= – 10(1) -2 (-1) + 4 (2)

= – 10 + 2 + 8

= 0

$\boxed{\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors}$
$\therefore\ The\ three\ vectors\ are\ mutually\ perpendicular.$

$\underline{Projection\ of\ \overrightarrow{a}\ on\ \overrightarrow{b}}$
$\color {purple} {Example\ 7\ .}\ \color {red} {Find\ the\ projection\ of\ the\ vector}\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}+2\overrightarrow{k} \ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}+2 \overrightarrow{j}+ 2\overrightarrow{k}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}$
$=\frac{(3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}+ 2\overrightarrow{k})}{\sqrt{(1)^2 + (2)^2 + (2)^2 }}$
$= \frac{3(1)+ 4(2)- 5(2)}{\sqrt{(1 + 4 + 4 }}$
$= \frac{3+ 8- 10}{\sqrt{9}}$
$\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{1}{3}}$

## Angle between two vectors using scalar product

$\underline{Angle\ between\ two\ vectors\ \overrightarrow{a} and\ \overrightarrow{b}}: \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$\color {purple} {Example\ 8\ .}\ \color {red} {Find\ the\ angle\ between\ vectors}\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} and\ 2\overrightarrow{i} -3\overrightarrow{j}- 5\overrightarrow{k}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k}) .(2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k})$

=   3 ( 2 ) + 4 ( – 3 ) – 2 ( – 5 )

=    6  – 12 + 10

= 4

$\boxed{ \overrightarrow{a}.\overrightarrow{b}= 4}$
$\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2 + (-2)^2 }=\sqrt{(9 + 16 +4 }=\sqrt{29}$
$\overrightarrow{|b|} = \sqrt{(2)^2 + (-5)^2 + (-3)^2 }=\sqrt{(4 + 25 +9 }=\sqrt{38}$
$\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$= \frac{4}{\sqrt{29}\sqrt{38}}$
$\boxed{\theta = \cos ^-1 ( \frac{4}{\sqrt{29}\sqrt{38}})}$

$\color {purple} {Example\ 9\ .}\ \color {red} {Find\ the\ projection\ of\ the\ vector}\ 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} on\ 7\overrightarrow{i}+ \overrightarrow{j}+2\overrightarrow{k}\ .\hspace{8cm}$$Also\ \color {red} {find\ the\ angle\ between\ them}\ \hspace {7cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{b}= 7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}) .(7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k})$

=3 ( 7 ) + 1 ( 1) – 2 ( 2 )

=    21 + 1 – 4

=   18.

$\boxed{ \overrightarrow{a}.\overrightarrow{b}= 18}$
$\overrightarrow{|a|} = \sqrt{(3)^2 + (1)^2 + (-2)^2 }=\sqrt{(9 + 1 +4 }=\sqrt{14}$
$\overrightarrow{|b|} = \sqrt{(7)^2 + (1)^2 + (2)^2 }=\sqrt{(49 + 1 + 4 }=\sqrt{54}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} = \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}} = \frac{18}{ \sqrt{54}}$
$\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$= \frac{18}{\sqrt{14}\sqrt{54}}$
$\boxed{\theta = \cos ^-1 ( \frac{18}{\sqrt{14}\sqrt{54}})}$

### APPLICATION OF SCALAR PRODUCT

#### Work done

$Work\ done = \overrightarrow{F}.\overrightarrow{d}\ where\ \overrightarrow{d}= \overrightarrow {OB}- \overrightarrow{OA}$
$\color {purple} {Example\ 10\ .}\ \color {red} {Find\ the\ work\ done}\ by\ the\ force\ 3\overrightarrow{i}+ 5\overrightarrow{j}+ 7\overrightarrow{k},\ \hspace{8cm}$$when\ the\ displacement\ is\ 2\overrightarrow{i}- \overrightarrow{j} +\overrightarrow{k}\ \hspace{7cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{F}= 3\overrightarrow{i}+ 5\overrightarrow{j}+7\overrightarrow{k}$
$\overrightarrow{d}= 2\overrightarrow{i}- \overrightarrow{j}+\overrightarrow{k}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}= (3\overrightarrow{i}+ 5\overrightarrow{j}+ 7\overrightarrow{k}) .(2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k})$

= 3 ( 2 ) + 5 ( – 1 ) + 7 ( 1 )

=    6  – 5  +  7

Work done = 8 units

$\boxed{Work\ done\ =\ 8\ units}$

$\color {purple} {Example\ 11\ .}\ A\ particle\ acted\ on\ by\ the\ forces\ 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} and\ \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k}\ \hspace{5cm}$$acting\ on\ the\ particle\ displaces\ the\ particle\ from\ the\ point\ \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k} to\ the\ point\ 3\overrightarrow{i}- 5\overrightarrow{j}+4\overrightarrow{k}\ .\ \hspace{5cm}$$\color {red} {Find\ the\ total\ work\ done\ by\ the\ forces}.\ \hspace {10cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$\overrightarrow{F_1}= 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{F_2}= \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k}$
$\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} + \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k}$
$\boxed{\overrightarrow{F}= 4\overrightarrow{i}+ 9\overrightarrow{j}+4\overrightarrow{k}}$
$\overrightarrow{OA}= \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}-5\overrightarrow{j}+ 4\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=3\overrightarrow{i}\ – 5\overrightarrow{j} + 4\overrightarrow{k}- (\overrightarrow{i}\ + 2\overrightarrow{j}+3\overrightarrow{k})$
$=\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k}- \overrightarrow{i}\ – 2\overrightarrow{j}- 3\overrightarrow{k}$
$\boxed{\overrightarrow{d}= 2\overrightarrow{i}- 7\overrightarrow{j}+\overrightarrow{k}}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}= (4\overrightarrow{i}+ 9\overrightarrow{j}+ 4\overrightarrow{k}) .(2\overrightarrow{i}-7 \overrightarrow{j}+ 1\overrightarrow{k})$

= 4 ( 2 ) + 9 ( – 7 ) + 4 ( 1 )

=    8  – 63  +  4

Work done  =  –51 units

$\boxed{Work\ done\ =\ 51\ units}\ (by\ taking\ positive\ value)$

$\color {purple} {Example\ 12\ .}A\ particle\ acted\ on\ by\ the\ forces\ 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} and\ 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\ \hspace{5cm}$$is\ displaced\ from\ the\ point\ ( 1, 1, 1 )\ to\ the\ point\ ( 2, – 3, 5 ).$$\color {red} {Find\ the\ total\ work\ done}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$\overrightarrow{F_1}= 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{F_2}= 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}$
$\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} + 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}$
$\boxed{\overrightarrow{F}= 6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}}$
$\overrightarrow{OA}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}-3\overrightarrow{j}+ 5\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=2\overrightarrow{i}\ – 3\overrightarrow{j} + 5\overrightarrow{k}- (\overrightarrow{i}\ + \overrightarrow{j}+\overrightarrow{k})$
$=2\overrightarrow{i}\ – 3\overrightarrow{j} + 5\overrightarrow{k}- \overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}$
$\boxed{\overrightarrow{d}= \overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k}}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}= (6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}) .(\overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k})$

=  6 ( 1 )  + 10 ( -4 ) – 1 ( 4 )

=    6  – 40  –  4

=    – 38

Work done    =  –38 units

$\boxed{Work\ done\ =\ 38\ units}\ (by\ taking\ positive\ value)$

$\color {purple} {Example\ 13\ .}A\ particle\ is\ displaced\ from\ the\ point\ 5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k}\ to\ the\ point\ 6\overrightarrow{i}+ 2\overrightarrow{j}-2\overrightarrow{k}$$under\ the\ action\ of\ forces\ 10\overrightarrow{i}- \overrightarrow{j} + 11\overrightarrow{k},\ 4\overrightarrow{i} + 5\overrightarrow{j} +6\overrightarrow{k}\ and\ -2\overrightarrow{i} + \overrightarrow{j}- 9\overrightarrow{k}.$$\color {red} {Calculate\ the\ total\ work\ done\ by\ the\ forces}.\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$\overrightarrow{F_1}= 10\overrightarrow{i}- \overrightarrow{j}+ 11\overrightarrow{k}$
$\overrightarrow{F_2}= 4\overrightarrow{i} + 5\overrightarrow{j}+ 6\overrightarrow{k}$
$\overrightarrow{F_3}= -2\overrightarrow{i}+ \overrightarrow{j}- 9\overrightarrow{k}$
$\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3}$$= 10\overrightarrow{i}- \overrightarrow{j}+ 11\overrightarrow{k} + 4\overrightarrow{i} + 5\overrightarrow{j}+ 6\overrightarrow{k}- 2\overrightarrow{i} + \overrightarrow{j}- 9\overrightarrow{k}$
$\boxed{\overrightarrow{F}= 12\overrightarrow{i}+ 5\overrightarrow{j} +8\overrightarrow{k}}$
$\overrightarrow{OA}= 5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k}$
$\overrightarrow{OB}= 6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}- (5\overrightarrow{i}- 5\overrightarrow{j}- 7\overrightarrow{k})$
$=6\overrightarrow{i} +2\overrightarrow{j} – 2\overrightarrow{k}- 5\overrightarrow{i} + 5\overrightarrow{j}+ 7\overrightarrow{k}$
$\boxed{\overrightarrow{d}= \overrightarrow{i}+ 7\overrightarrow{j}+5\overrightarrow{k}}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}= (12\overrightarrow{i}+ 5\overrightarrow{j} +8\overrightarrow{k}) .(\overrightarrow{i}+ 7\overrightarrow{j}+5\overrightarrow{k})$
$=12(1) + 5(7) + 8(5)$
$= 12 + 35 + 40$
$\boxed{Work\ done\ = 87\ units}$

### Vector Product of Two Vectors

#### Definition

$Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.$ $Then\ the\ vector\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by$$\overrightarrow{a}×\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}×\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\sin\theta\ n^\wedge$

Properties of Vector Product:

$1. \overrightarrow{a}\ and\overrightarrow{b}are\ parellel\ vectors\ if\ and\ only\ if \overrightarrow{a}× \overrightarrow{b}= 0.$
$2.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ any\ two\ vectors\ then \overrightarrow{a}×\overrightarrow{b} = -\overrightarrow{b}×\overrightarrow{a}$
$3. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively$
$Then\ i) \overrightarrow{i}×\overrightarrow{i}= \overrightarrow{j}×\overrightarrow{j}=\overrightarrow{k}×\overrightarrow{k}=0$
$ii) \overrightarrow{i}×\overrightarrow{j}= \overrightarrow{k};\overrightarrow{j}×\overrightarrow{k}= \overrightarrow{i};\overrightarrow{k}×\overrightarrow{i}=\overrightarrow{j}$
$iii) \overrightarrow{j}×\overrightarrow{i}= -\overrightarrow{k};\overrightarrow{k}×\overrightarrow{j}= -\overrightarrow{i};\overrightarrow{i}×\overrightarrow{k}= -\overrightarrow{j}$

4. Vector product in determinant from

$Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \overrightarrow{a_1} & \overrightarrow{a_2} & \overrightarrow{a_3}\\ \overrightarrow{b_1} & \overrightarrow{b_2} & \overrightarrow{b_3}\\ \end{vmatrix}$
$5.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|$
$6.\ If \overrightarrow{d_1}\ and \overrightarrow{d_2}\ are\ two\ diagonals\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = \frac{1}{2}|\overrightarrow{d_1} × \overrightarrow{d_2}|$
$7.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ triangle. Then\ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{a} × \overrightarrow{b}|$
$8.\ Area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC}\ is\ \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|$
$9.\ \ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$10.\ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}$
$\color {purple} {Example\ 14\ .}\ \color {red} {Find\ \overrightarrow{a}× \overrightarrow{b}}\ if\ \overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}+ 3\overrightarrow{k}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$\overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i}-\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 1\\ 2 & -1 & 3\\ \end{vmatrix}$
$= \overrightarrow{i}( 3 + 1) -\overrightarrow{j}(3-2)+\overrightarrow{k}(-1-2)$
$= \overrightarrow{i}(4) -\overrightarrow{j}(1)+\overrightarrow{k}(-3)$
$\boxed{ \overrightarrow{a}× \overrightarrow{b}= 4\overrightarrow{i}-\overrightarrow{j}-3\overrightarrow{k}}$

$\color {purple} {Example\ 15\ .}\ \color {red} {Show\ that}\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 4\overrightarrow{k}\ and\ 3\overrightarrow{i}\ -\ 6\overrightarrow{j}\ -\ 12\overrightarrow{k}\ \color {red} {are\ parallel}.\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$To\ show\ \overrightarrow{a}×\overrightarrow{b} =\ 0\ \hspace{10cm}$
$\overrightarrow{a}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 4\overrightarrow{k}$
$\overrightarrow{b}\ =\ 3\overrightarrow{I}\ -\ 6\overrightarrow{j}\ -\ 12\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & -2 & -4\\ 3 & -6 & -12\\ \end{vmatrix}$
$= \overrightarrow{i}( 24\ -\ 24)\ -\overrightarrow{j}(-12\ +\ 12)\ +\ \overrightarrow{k}(-6\ +\ 6)$
$= \overrightarrow{i}(0) -\overrightarrow{j}(0)+\overrightarrow{k}(0)$
$\boxed{\overrightarrow{a}×\overrightarrow{b} =\ 0}\ \hspace{7cm}$
$\therefore\ The\ given\ vectors\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ parallel$

$\color {purple} {Example\ 16\ .}\ \color {red} {Find\ the\ area\ of\ the\ parellelogram}\ whose\ adjacent\ sides\ are\ \hspace{10cm}$$-\ \overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\ and\ \overrightarrow{i}\ – \overrightarrow{j}\ -\ \overrightarrow{k}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}=-\ \overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}\ – \overrightarrow{j}\ -\ \overrightarrow{k}$
$Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ – 1 & 2 & 4\\ 1 & -1 & -1\\ \end{vmatrix}$
$= \overrightarrow{i}(-\ 2 +\ 4)\ -\ \overrightarrow{j}(1\ -\ 4)\ +\ \overrightarrow{k}(1\ -\ 2)$
$= \overrightarrow{i}(2)\ -\overrightarrow{j}(-\ 3)\ +\ \overrightarrow{k}(- 1)$
$\overrightarrow{a}× \overrightarrow{b}= 2\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(2)^2\ +\ (3)^2 + (-1)^2 }=\sqrt{(4\ +\ 9\ +\ 1}=\sqrt{14}$
$\boxed{Area\ of \ parellelogram = \sqrt{14} sq.units}$

$\color {purple} {Example\ 17\ .}\ \color {red} {Find\ the\ area\ of\ the\ parellelogram}\ whose\ diagonals\ are\ represented\ by\ \hspace{10cm}$$3\ \overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k}\ and\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\ \hspace{5cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\color {purple} {Example\ 18\ .}\ \color {red} {Find\ the\ area\ of\ the\ triangle}\ formed\ by\ the\ points\ whose\ position\ vectors\ \hspace{8cm}$$3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\overrightarrow{k}\ , 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ \overrightarrow{k}\ and\ 5\overrightarrow{i}\ + \overrightarrow{j}\ +\ 3\overrightarrow{k}\ \hspace{7cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{OA}= 3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}$
$\overrightarrow{OC}= 5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- (3\overrightarrow{i} + 2\overrightarrow{j}- \overrightarrow{k})$
$=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- 3\overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}$
$\boxed{\overrightarrow{AB}= -\overrightarrow{i} – 5\overrightarrow{j} + 2\overrightarrow{k}}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- (2\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})$
$=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- 2\overrightarrow{i} + 3\overrightarrow{j} -\overrightarrow{k}$
$\boxed{\overrightarrow{BC}= 3\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}}$
$\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -1 & -5 & 2\\ 3 & 4 & 2\\ \end{vmatrix}$
$= \overrightarrow{i}( -10 – 8) -\overrightarrow{j}(-2 – 6)+\overrightarrow{k}(-4 + 15)$
$= \overrightarrow{i}(-18) -\overrightarrow{j}(8)+\overrightarrow{k}(11)$
$\boxed{\overrightarrow{AB}× \overrightarrow{BC}= -18\overrightarrow{i}\ +\ 8\overrightarrow{j}+11\overrightarrow{k}}$
$|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-18 )^2 + (-8)^2 + (11)^2 }=\sqrt{(324 + 64 + 121 }=\sqrt{509}$
$Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|$
$\boxed{Area\ of \ triangle\ =\frac{\sqrt{509}}{2}\ sq. units}$

$\color {purple} {Example\ 19\ .}\ If\ |\overrightarrow{a}| = 2,\ |\overrightarrow{b}|= 7\ and\ |\overrightarrow{a} × \overrightarrow{b}|=7,\ \color {red} {find\ the\ angle\ between\ \overrightarrow{a}\ and\ \overrightarrow{b}}.\ \hspace {3cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$=\frac{7}{(2)(7)} = \frac{1}{2}$
$\boxed{\theta = 30^{\circ}}$

$\color {purple} {Example\ 20\ .}\ \color {red} {Find\ the\ unit\ vector}\ perpendicular\ to\ each\ of\ the\ vectors\ 2\overrightarrow{i} -\overrightarrow{j}+\overrightarrow{k} and\ 3\overrightarrow{i}+ 4\overrightarrow{j} -\overrightarrow{k}.$$\color {red} {Also\ find\ the\ sine\ of\ the\ angle}\ between\ the\ vectors .$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i} -\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{b}=3\overrightarrow{i}+ 4\overrightarrow{j} -\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 2 & -1 & 1\\ 3 & 4 & -1\\ \end{vmatrix}$
$= \overrightarrow{i}( 1 – 4) -\overrightarrow{j}(-2 – 3)+\overrightarrow{k}(8 + 3)$
$= \overrightarrow{i}(-3) -\overrightarrow{j}(-5)+\overrightarrow{k}(11)$
$\boxed{\overrightarrow{a}× \overrightarrow{b}= -3\overrightarrow{i} +5\overrightarrow{j}+11\overrightarrow{k}}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-3 )^2 + (5)^2 + (11)^2 }=\sqrt{9 + 25 + 121}=\sqrt{155}$
$n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{-3\overrightarrow{i}\ + 5\overrightarrow{j}+ 11\overrightarrow{k}}{\sqrt{155}}$
$\boxed{n^\wedge = \frac{-3\overrightarrow{i}\ + 5\overrightarrow{j}+ 11\overrightarrow{k}}{\sqrt{155}}}$
$\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2 + (1)^2 }=\sqrt{(4 + 1 + 1 }=\sqrt{6}$
$\overrightarrow{|b|} = \sqrt{(3)^2 + (4)^2 + (-1)^2 }=\sqrt{(9 + 16 + 1 }=\sqrt{26}$
$\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{155}}{\sqrt{6}\sqrt{26}}$
$\boxed{sin\ \theta =\frac{\sqrt{155}}{\sqrt{6}\sqrt{26}}}$

### Application of Vector Product

Moment (or) Torque of a force about a point

$Let\ A\ be\ any\ point\ and\ \overrightarrow{r}\ be\ the\ position\ vector\ relative\ to\ the\ point\ A$$of\ any\ point\ P\ on\ the\ line\ of\ the\ action\ of\ the\ force\ \overrightarrow{F}.$$The\ moment\ of\ the\ force\ about\ the\ point\ A\ is\ defined\ as\ \overrightarrow{M}= \overrightarrow{r}× \overrightarrow{F}$$where\ \overrightarrow{r}= \overrightarrow{AP}= \overrightarrow{OP}- \overrightarrow{OA}$.
$The\ Magnitude \ of\ Moment = |\overrightarrow{r} × \overrightarrow{F}|$
$\color {purple} {Example\ 21\ .}\ \color {red} {Find\ the\ moment\ of\ the\ force}\ 3\overrightarrow{i}\ +\ \overrightarrow{k}\ \hspace{8cm}$$acting\ through\ the\ point\ \overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}\ about\ the\ point\ 2\overrightarrow{i}+ \overrightarrow{j}-2\overrightarrow{k}.$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{F}= 3\overrightarrow{i} + \overrightarrow{k}$
$\overrightarrow{OP}= \overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}$
$\overrightarrow{OA}= 2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}$
$\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}$
$=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- (2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k})$
$=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- 2\overrightarrow{i}-\overrightarrow{j} +2\overrightarrow{k})$
$\boxed{\overrightarrow{r}= -\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}}$
$Moment = \overrightarrow{r}× \overrightarrow{F}$
$=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -1 & 1 & 1\\ 3 & 0 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 1 – 0) -\overrightarrow{j}(-1 – 3)+\overrightarrow{k}(0 – 3)$
$= \overrightarrow{i}(1) -\overrightarrow{j}(-4)+\overrightarrow{k}(-3)$
$\boxed{\overrightarrow{r}× \overrightarrow{F}= \overrightarrow{i}+ 4\overrightarrow{j}-3\overrightarrow{k}}$
$Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|$
$= \sqrt{(1)^2 + (4)^2 + (-3)^2 }=\sqrt{(1 + 16 +9 }=\sqrt{26}$
$\boxed{Magnitude\ of \ Moment = \sqrt{26}\ units}$

$\color {purple} {Example\ 22\ .}\ \color {red} {Find\ the\ Magnitude\ of\ the\ moment\ of\ the\ force}\ 6\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}\ \hspace{7cm}$$acting\ along\ the\ point\ (0,\ 1,\ -\ 1)\ about\ the\ point\ (4,\ 3,\ -\ 1)\ \hspace{8cm}$
$\color {blue} {Soln:}\ \hspace{18cm}$
$\overrightarrow{F}= 6\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OA}= 4\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OP}\ =\ 0\overrightarrow{i}\ + \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OP}\ =\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}$
$=\overrightarrow{j}\ -\ \overrightarrow{k}\ – (4\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k})$
$=\overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 4\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ \overrightarrow{k}$
$\boxed{\overrightarrow{r}\ =\ -\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}}$
$Moment = \overrightarrow{r}× \overrightarrow{F}$
$=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -4 & -2 & 0\\ 6 & 1 & – 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 2 – 0) -\overrightarrow{j}(4 – 0)\ +\ \overrightarrow{k}(-4 + 12)$
$Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|$
$= \sqrt{(2)^2 + (-4)^2 + (8)^2 }=\sqrt{(4\ +\ 16\ +\ 64 }\ =\sqrt{84}$
$\boxed{Magnitude\ of \ Moment = \sqrt{84}}$