2.3 APPLICATION OF SCALAR AND VECTOR PRODUCT

Application of Scalar Product

Work done

Multiplying vectors together | Work physics, Physics, Angles math
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}\ where\ \overrightarrow{d}= \overrightarrow {OB}- \overrightarrow{OA}\]

Example  :

\[A\ particle\ acted\ on\ by\ the\ forces\ 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} and\ \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} acting\ on\ the\ particle\]
\[displaces\ the\ particle\ from\ the\ point\ \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k} to\ the\ point\ 3\overrightarrow{i}- 5\overrightarrow{j}+4\overrightarrow{k}. Find\ the\ total\ work\ done\ by\ the\ forces.\]

Soln:

\[\overrightarrow{F_1}= 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{F_2}= \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} \]
\[\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} + \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k} \]
\[\overrightarrow{F}= 4\overrightarrow{i}+ 9\overrightarrow{j}+4\overrightarrow{k} \]
\[\overrightarrow{OA}= \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k} \]
\[\overrightarrow{OB}= 3\overrightarrow{i}-5\overrightarrow{j}+ 4\overrightarrow{k} \]
\[\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}\]
\[=3\overrightarrow{i}\ - 5\overrightarrow{j} + 4\overrightarrow{k}- (\overrightarrow{i}\ + 2\overrightarrow{j}+3\overrightarrow{k})\]
\[=\overrightarrow{i}\ - \overrightarrow{j} + 3\overrightarrow{k}- \overrightarrow{i}\ - 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{d}= 2\overrightarrow{i}- 7\overrightarrow{j}+\overrightarrow{k} \]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (4\overrightarrow{i}+ 9\overrightarrow{j}+ 4\overrightarrow{k}) .(2\overrightarrow{i}-7 \overrightarrow{j}+ 1\overrightarrow{k})\]

= 4 ( 2 ) + 9 ( - 7 ) + 4 ( 1 )

=    8  - 63  +  4

    Work done  =  --51 units

Work done  =  51 units  (by taking positive value)

Example  :

\[A\ particle\ acted\ on\ by\ the\ forces\ 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} and\ 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]

is displaced from the point ( 1, 1,  1 )  to the  point  ( 2, - 3, 5 ).  Find the total work done.

Soln:

\[\overrightarrow{F_1}= 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{F_2}= 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]
\[\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 4\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k} + 2\overrightarrow{i}+ 7\overrightarrow{j}-2\overrightarrow{k}\]
\[\overrightarrow{F}= 6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}\]
\[\overrightarrow{OA}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}-3\overrightarrow{j}+ 5\overrightarrow{k}\]
\[\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}\]
\[=2\overrightarrow{i}\ - 3\overrightarrow{j} + 5\overrightarrow{k}- (\overrightarrow{i}\ + \overrightarrow{j}+\overrightarrow{k})\]
\[=2\overrightarrow{i}\ - 3\overrightarrow{j} + 5\overrightarrow{k}- \overrightarrow{i}\ - \overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{d}= \overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k}\]
\[Work\ done = \overrightarrow{F}.\overrightarrow{d}= (6\overrightarrow{i}+ 10\overrightarrow{j}-\overrightarrow{k}) .(\overrightarrow{i}- 4\overrightarrow{j}+4\overrightarrow{k})\]

    =  6 ( 1 )  + 10 ( -4 ) - 1 ( 4 )

=    6  - 40  -  4

=    - 38

Work done    =  --38 units

Work done    =  38 units  (by taking positive value)

Application of Vector Product

Moment (or) Torque of a force about a point

Torque or Moment of a Force - QS Study
\[Let\ A\ be\ any\ point\ and\ \overrightarrow{r}\ be\ the\ position\ vector\ relative\ to\ the\ point\ A\]\[of\ any\ point\ P\ on\ the\ line\ of\ the\ action\ of\ the\ force\ \overrightarrow{F}.\]\[The\ moment\ of\ the\ force\ about\ the\ point\ O\ is\ defined\ as\ \overrightarrow{M}= \overrightarrow{r}× \overrightarrow{F}\]\[where\ \overrightarrow{r}= \overrightarrow{AP}= \overrightarrow{OP}- \overrightarrow{OA}\].
\[ The\ Magnitude \ of\ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]

Example  :

\[Find\ the\ moment\ of\ the\ force\ 3\overrightarrow{i}+\overrightarrow{k}\ acting\ through\ the\ point\ \overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k} about\ the\ point\ 2\overrightarrow{i}+ \overrightarrow{j}-2\overrightarrow{k}.\]

Soln:

\[\overrightarrow{F}= 3\overrightarrow{i} + \overrightarrow{k}\]
\[\overrightarrow{OP}= \overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}\]
\[\overrightarrow{OA}= 2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}\]
\[\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}\]
\[=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- (2\overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k})\]
\[=\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}- 2\overrightarrow{i}-\overrightarrow{j} +2\overrightarrow{k})\]
\[\overrightarrow{r}= -\overrightarrow{i} + \overrightarrow{j} + \overrightarrow{k}\]
\[Moment = \overrightarrow{r}× \overrightarrow{F}\]
\[ =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -1 & 1 & 1\\ 3 & 0 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 1 - 0) -\overrightarrow{j}(-1 - 3)+\overrightarrow{k}(0 - 3)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(-4)+\overrightarrow{k}(-3)\]
\[\overrightarrow{r}× \overrightarrow{F}= \overrightarrow{i}+ 4\overrightarrow{j}-3\overrightarrow{k}\]
\[ Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
\[= \sqrt{(1)^2 + (4)^2 + (-3)^2 }=\sqrt{(1 + 16 +9 }=\sqrt{26}\]
\[ Magnitude\ of \ Moment = \sqrt{26}\ units\]

Example  :

\[Find\ the\ moment\ of\ the\ force\ 3\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}\ acting\ through\ the\ point\ \overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k} about\ the\ point\ 4\overrightarrow{i}- 3\overrightarrow{j}+\overrightarrow{k}.\]

Soln:

\[\overrightarrow{F}= 3\overrightarrow{i} +4\overrightarrow{j} + 5\overrightarrow{k}\]
\[\overrightarrow{OP}= \overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{OA}= 4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k}\]
\[\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}\]
\[=\overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k}- (4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})\]
\[=\overrightarrow{i} - 2\overrightarrow{j} + 3\overrightarrow{k}- 4\overrightarrow{i} +3\overrightarrow{j} -\overrightarrow{k})\]
\[\overrightarrow{r}= -3\overrightarrow{i} + \overrightarrow{j} + 2\overrightarrow{k}\]
\[Moment = \overrightarrow{r}× \overrightarrow{F}\]
\[ =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -3 & 1 & 2\\ 3 & 4 & 5\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 5 - 8) -\overrightarrow{j}(-15 - 6)+\overrightarrow{k}(-12 - 3)\]
\[\overrightarrow{r}× \overrightarrow{F}= -3\overrightarrow{i}+ 21\overrightarrow{j}-15\overrightarrow{k}\]
\[ Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
\[= \sqrt{(-3)^2 + (21)^2 + (-15)^2 }=\sqrt{(9 + 441+ 225 }=\sqrt{675}\]
\[ Magnitude\ of \ Moment = \sqrt{675}\ units\]

2.2 PRODUCT OF VECTORS

Multiplication of vectors -Study Material for IIT JEE | askIITians
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:
\[1. \overrightarrow{a}\ and\overrightarrow{b}are\ perpendicular\ vectors\ if\ and\ only\ if \overrightarrow{a}.\overrightarrow{b}=0. \]
\[2. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively \]
\[Then\ i) \overrightarrow{i}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1 \]
\[ ii) \overrightarrow{i}.\overrightarrow{j}= \overrightarrow{j}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{i}= \overrightarrow{k}.\overrightarrow{j}=\overrightarrow{i}.\overrightarrow{k}=0 \]
\[3. \ Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[4. \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[If\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[ \overrightarrow{a}.\overrightarrow{b}= a_1b_1 + a_2b_2 + a_3b_3\ (using\ property\ 2)\]

Example  :

\[Find\ the\ scalar\ product\ of\ \overrightarrow{i}+\overrightarrow{j}\ ,\ \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= \overrightarrow{i} + \overrightarrow{j} \]
\[\overrightarrow{b}= \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+\overrightarrow{j}) .(\overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k})\]

= 1(1) + 1(1) + 0

= 1 + 1

= 2

\[ \overrightarrow{a}.\overrightarrow{b}= 2\]

Example  :

\[Show\ that\ the\ vectors\ 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k} and\ 3\overrightarrow{i}+ 2\overrightarrow{j}+6\overrightarrow{k} are\ perpendicular\ to\ each\ other \]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k}) .(3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k})\]

= 2(3) + 3(2) – 2 (6)

= 6 + 6 -12

= 0

\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[2.\ Find\ the\ value\ of\ m\ if\ the\ vectors\ 2\overrightarrow{i}+ m\overrightarrow{j}- 3\overrightarrow{k} and\ 3\overrightarrow{i}+ 1\overrightarrow{j}+4\overrightarrow{k} are\ perpendicular\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ m\overrightarrow{j}- 3\overrightarrow{k} \]
\[ \overrightarrow{b}= 3\overrightarrow{i}+ 1\overrightarrow{j}+4\overrightarrow{k} \]
\[Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other \]
\[i.e\ \overrightarrow{a}.\overrightarrow{b}= 0\]
\[(2\overrightarrow{i} + m\overrightarrow{j}-3k).(3\overrightarrow{i}+ \overrightarrow{j}+4\overrightarrow{k}) = 0 .\]

2(3) + m(1) – 3 (4) = 0

6 + m -12 = 0

m – 6 = 0

m = 6.

\[3.\ Prove\ that\ the\ vectors\ \overrightarrow{i}-\overrightarrow{j}+ 2\overrightarrow{k},\ \overrightarrow4{j}+ 2\overrightarrow{k}and\ -10\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k} are\ mutually\ perpendicular.\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k} \]
\[ \overrightarrow{b}= + 4\overrightarrow{j}+2\overrightarrow{k} \]
\[\overrightarrow{c}= -10\overrightarrow{i}- 2\overrightarrow{j}+ 4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k}) .(4\overrightarrow{j}+ 2\overrightarrow{k})\]

= 1(0) – 1(4) + 2 (2)

= 0

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

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

= 0 – 8 + 8

= 0

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

= – 10(1) -2 (-1) + 4 (2)

= – 10 + 2 + 8

= 0

\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]

\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b}\]

Example  :

\[Find\ the\ projection\ of\ the\ vector\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}+2\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}+2 \overrightarrow{j}+ 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}+ 2\overrightarrow{k})}{\sqrt{(1)^2 + (2)^2 + (2)^2 }}\]
\[= \frac{3(1)+ 4(2)- 5(2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{3+ 8- 10}{\sqrt{9}}\]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{1}{3}\]
\[Angle\ between\ two\ vectors\ \overrightarrow{a} and\ \overrightarrow{b}: \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]

Example  :

\[Find\ the\ angle\ between\ vectors\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} and\ 2\overrightarrow{i} -3\overrightarrow{j}- 5\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k}) .(2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k})\]

=   3 ( 2 ) + 4 ( – 3 ) – 2 ( – 5 )

=    6  – 12 + 10

= 4

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

Example  :

\[Find\ the\ projection\ of\ the\ vector\ 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} on\ 7\overrightarrow{i}+ \overrightarrow{j}+2\overrightarrow{k}. Also\ find\ the\ angle\ between\ them \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}) .(7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k})\]

=3 ( 7 ) + 1 ( 1) – 2 ( 2 ) 

   =    21 + 1 – 4

=   18.

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

Vector Product of Two Vectors

Definition

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

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

\[Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \overrightarrow{a_1} & \overrightarrow{a_2} & \overrightarrow{a_3}\\ \overrightarrow{b_1} & \overrightarrow{b_2} & \overrightarrow{b_3}\\ \end{vmatrix}\]
\[5.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[6.\ If \overrightarrow{d_1}\ and \overrightarrow{d_2}\ are\ two\ diagonals\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = \frac{1}{2}|\overrightarrow{d_1} × \overrightarrow{d_2}|\]
\[7.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ triangle. Then\ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{a} × \overrightarrow{b}|\]
\[8.\ Area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC}\ is\ \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[9.\ \ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[10.\ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}\]

Example  :

\[Find\ \overrightarrow{a}× \overrightarrow{b}if\ \overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}+ 3\overrightarrow{k}\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}-\overrightarrow{j}+ 3\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 1\\ 2 & -1 & 3\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 3 + 1) -\overrightarrow{j}(3-2)+\overrightarrow{k}(-1-2)\]
\[ = \overrightarrow{i}(4) -\overrightarrow{j}(1)+\overrightarrow{k}(-3)\]
\[ \overrightarrow{a}× \overrightarrow{b}= 4\overrightarrow{i}-\overrightarrow{j}-3\overrightarrow{k}\]

Example  :

\[Find\ the\ area\ of\ the\ parellelogram\ whose\ adjacent\ sides\ are\ 3\overrightarrow{i}- \overrightarrow{k} and\ \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k}.\]

Soln:

\[\overrightarrow{a}=3\overrightarrow{i} – \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ 3 & 0 &-1\\ 1 & 1 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 0 + 1) -\overrightarrow{j}(3+1)+\overrightarrow{k}(3-0)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(4)+\overrightarrow{k}(3)\]
\[ \overrightarrow{a}× \overrightarrow{b}= \overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(1)^2 + (-4)^2 + (3)^2 }=\sqrt{(1 + 16 +9 }=\sqrt{26}\]
\[ Area\ of \ parellelogram = \sqrt{26} sq.units\]


Example  :

\[Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ 3\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}\, 2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k} and\ 5\overrightarrow{i}+ \overrightarrow{j}+3\overrightarrow{k}.\]

Soln:

\[\overrightarrow{OA}= 3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}\]
\[\overrightarrow{OC}= 5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- (3\overrightarrow{i} + 2\overrightarrow{j}- \overrightarrow{k})\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- 3\overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{AB}= -\overrightarrow{i} – 5\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- (2\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- 2\overrightarrow{i} + 3\overrightarrow{j} -\overrightarrow{k})\]
\[\overrightarrow{BC}= 3\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -1 & -5 & 2\\ 3 & 4 & 2\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -10 – 8) -\overrightarrow{j}(-2 – 6)+\overrightarrow{k}(-4 + 15)\]
\[ = \overrightarrow{i}(-18) -\overrightarrow{j}(-8)+\overrightarrow{k}(11)\]
\[\overrightarrow{AB}× \overrightarrow{BC}= -18\overrightarrow{i}-8\overrightarrow{j}+11\overrightarrow{k}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-18 )^2 + (-8)^2 + (11)^2 }=\sqrt{(324 + 64 + 121 }=\sqrt{509}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[=\frac{\sqrt{509}}{2}\ sq. units\]

Example  :

\[ If\ |\overrightarrow{a}| = 2,\ |\overrightarrow{b}|= 7\ and\ |\overrightarrow{a} × \overrightarrow{b}|=7,\ find\ the\ angle\ between\ \overrightarrow{a}\ and\ \overrightarrow{b}.\]

Soln:

\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[ =\frac{7}{(2)(7)} = \frac{1}{2}\]
\[\theta = 30^{\circ} \]

Example  :

\[Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ 2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} and\ \overrightarrow{i}+ 2\overrightarrow{j}+\overrightarrow{k}.\]\[Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ 2 & 1 & 1\\ 1 & 2 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 1 – 2) -\overrightarrow{j}(2 – 1)+\overrightarrow{k}(4 – 1)\]
\[ = \overrightarrow{i}(-1) -\overrightarrow{j}(1)+\overrightarrow{k}(3)\]
\[\overrightarrow{a}× \overrightarrow{b}= -\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-1 )^2 + (-1)^2 + (3)^2 }=\sqrt{(1 + 1 + 9}=\sqrt{11}\]
\[ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{\overrightarrow{i}\ – \overrightarrow{j}+ 3\overrightarrow{k}}{\sqrt{11}} \]
\[ n^\wedge = \frac{\overrightarrow{i}\ – \overrightarrow{j}+ 3\overrightarrow{k}}{\sqrt{11}} \]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (1)^2 + (1)^2 }=\sqrt{(4 + 1 + 1 }=\sqrt{6}\]
\[\overrightarrow{|b|} = \sqrt{(1)^2 + (2)^2 + (1)^2 }=\sqrt{(1 + 4 + 1 }=\sqrt{6}\]
\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{11}}{\sqrt{6}\sqrt{6}} = \frac{\sqrt{11}}{6}\]
\[\ sin\ \theta =\frac{\sqrt{11}}{6}\]

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:

Addition of Vectors - Study Page
\[\overrightarrow{OA} = \overrightarrow{a}\] \[\overrightarrow{AB} = \overrightarrow{b}\]\[\overrightarrow{OA}+ \overrightarrow{AB} = \overrightarrow{OB}\]\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]

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 Cosines and 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.

Direction Cosines & Direction Ratios - Definitions & Examples
\[\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}\]
\[OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }= r\]
\[ Direction\ cosines\ are \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \]

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.