# N – 2.2 – Product of two vectors – Exercise Problems with solutions

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} { What\ are\ the\ values\ of}\ \overrightarrow{i}\ . \ \overrightarrow{j}\ ,\ and\ \overrightarrow{k}\ .\ \overrightarrow{k}\ \hspace{18cm}$
$\color {blue} {Soln:}\ \overrightarrow{i}\ . \ \overrightarrow{j}\ =\ 0\ and\ \overrightarrow{k}\ .\ \overrightarrow{k}\ =\ 0\ \hspace{15cm}$
$\color {purple} {2\ .}\ \color {red} {Find\ the\ scalar\ product\ of}\ \overrightarrow{i}+\overrightarrow{j}\ ,\ \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k}\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= \overrightarrow{i} + \overrightarrow{j}$
$\overrightarrow{b}= \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+\overrightarrow{j}) .(\overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k})$

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

= 1 + 1

= 2

$\overrightarrow{a}.\overrightarrow{b}= 2$

$\color {purple} {3\ .}\ \color {red} {Show\ that\ the\ vectors}\ \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}\ and\ – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k}\ are\ perpendicular\ to\ each\ other\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}$
$\overrightarrow{b}= – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}) .(- 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k})$

= 2(3) + 3(2) – 2 (6)

= 6 + 6 -12

= 0

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

$\color {purple} {4\ .}\ \color {red} {Find\ the\ value\ of\ p}\ if\ the\ vectors\ 2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k}\ and\ p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k} are\ perpendicular.\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k}$
$\overrightarrow{b}= p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k}$
$Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other$
$i.e\ \overrightarrow{a}.\overrightarrow{b}= 0$
$(2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k} ).(p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k}) = 0 .$

2(p) + 1(3) – 5 (-2) = 0

2p + 3 +10 = 0

2p + 13 = 0

$p = \frac{-13}{2}$

$\color {purple} {5\ .}\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ two\ vectors\ such\ that\ |\overrightarrow{a}| = 6,\ |\overrightarrow{b}|= 4\ and\ \overrightarrow{a}\ . \overrightarrow{b}\ =12\ \hspace{5cm}$$\color {red} {find\ the\ angle\ between\ them.}\ \hspace{3cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$Given\ |\overrightarrow{a}| = 6,\ |\overrightarrow{b}|= 4\ and\ \overrightarrow{a}\ . \overrightarrow{b}\ =12\ \hspace{10cm}$
$\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$=\frac{12}{(6)(4)} = \frac{1}{2}$
$\theta\ =\ cos^{1}(\frac{1}{2})\ =\ 60^0\ =\ \frac{\pi}{3}$

$\color {purple} {6\ .}\ \color {red} {What\ are\ the\ values\ of}\ (i)\ \overrightarrow{i}\ . \ \overrightarrow{i}\ ,\ and\ (ii)\ \overrightarrow{i}\ ×\ \overrightarrow{j}\ \hspace{18cm}$
$\color {blue} {Soln:}\ (i)\ \overrightarrow{i}\ . \ \overrightarrow{i}\ =\ i\ \hspace{18cm}$
$(ii)\ \overrightarrow{i}\ ×\ \overrightarrow{j}\ =\ \overrightarrow{k}\ \hspace{12cm}$
$\color {purple} {7\ .}\ \color {red} {Prove\ that\ the\ vectors}\ \overrightarrow{a}\ =\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ \color {red} {are\ parallel}\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$To\ show\ \overrightarrow{a}×\overrightarrow{b} =\ 0\ \hspace{10cm}$
$\overrightarrow{a}= 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i}-\overrightarrow{j}\ -\ 3\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 4 & -2 & -6\\ 2 & -1 & -3\\ \end{vmatrix}$
$= \overrightarrow{i}( 6\ -\ 6)\ -\overrightarrow{j}(-12\ +\ 12)\ +\ \overrightarrow{k}(-4\ +\ 4)$
$= \overrightarrow{i}(0) -\overrightarrow{j}(0)+\overrightarrow{k}(0)$
$\boxed{\overrightarrow{a}×\overrightarrow{b} =\ 0}\ \hspace{7cm}$
$\therefore\ The\ given\ vectors\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ parallel$

$\color {purple} {8\ .}\ If\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ parallelogram.\ \hspace{15cm}$$\color {red} {What\ is\ its\ area?}\ \hspace{12cm}$
$\color {blue} {Soln:}\ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\ \hspace{17cm}$
$\color {purple} {9\ .}\ If\ |\overrightarrow{a}| = 3,\ |\overrightarrow{b}|= 5\ and\ |\overrightarrow{a} × \overrightarrow{b}|=10,\ \color {red} {find\ the\ angle\ between\ \overrightarrow{a}\ and\ \overrightarrow{b}}\ \hspace{10cm}$
$\color {blue} {Soln:}\ \hspace{20cm}$
$\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$=\frac{10}{(3)(5)} = \frac{2}{3}$
$\boxed{\theta\ = sin^{-1}(\frac{2}{3})}$

$\underline{PART\ -\ B}$
$10.\ Find\ the\ projection\ of\ the\ vector\ 2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} on\ the\ vector\ \overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k}\ \hspace{10cm}$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{b}=\overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}$
$=\frac{(2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}).(\overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k})}{\sqrt{(1)^2 + (-2)^2 + (-2)^2 }}$
$= \frac{2(1)+ 1(-2)- 2(-2)}{\sqrt{(1 + 4 + 4 }}$
$= \frac{2 – 2 + 4}{\sqrt{9}}$
$\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{4}{3}}$

$11.\ Find\ the\ projection\ of\ the\ vector\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}\ on\ the\ vector\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ \hspace{10cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{b}=\overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}$
$=\frac{(2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}).(3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})}{\sqrt{(3)^2 + (-1)^2 + (1)^2 }}$
$= \frac{2(3)\ +\ 3(-\ 1)\ +\ 1(1)}{\sqrt{(9\ +\ 1\ +\ 1 }}$
$= \frac{6\ -\ 3\ +\ 1}{\sqrt{11}}$
$\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b}\ =\ \frac{4}{\sqrt{11}}}$

$12.\ Find\ the\ area\ of\ the\ parellelogram\ whose\ adjacent\ sides\ are\ \overrightarrow{i} + \overrightarrow{j} + 3\overrightarrow{k}\ and\ 2\overrightarrow{i} + \overrightarrow{j}+ 2\overrightarrow{k}.\ \hspace{10cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}=\overrightarrow{i} + \overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i} + \overrightarrow{j}+ 2\overrightarrow{k}$
$Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 3\\ 2 & 1 & 2\\ \end{vmatrix}$
$= \overrightarrow{i}( 2 -3) -\overrightarrow{j}(3+1)+\overrightarrow{k}(3-0)$
$= \overrightarrow{i}(-1) -\overrightarrow{j}(- 4)+\overrightarrow{k}(- 1)$
$\overrightarrow{a}× \overrightarrow{b}= – \overrightarrow{i} + 4\overrightarrow{j} -\overrightarrow{k}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-1)^2 + (4)^2 + (-1)^2 }=\sqrt{(1 + 16 +1 }=\sqrt{18}$
$\boxed{Area\ of \ parellelogram = 3\sqrt{2} sq.units}$

$13.\ If\ \overrightarrow{d_1}\ =\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ and\ \overrightarrow{d_2}\ =\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ are\ \hspace{15cm}$$diagonals\ of\ a\ parellelogram.\ Find\ its\ Area\ \hspace{10cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{d_1}\ =\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{d_1}\ =\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}\ =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 4 & 2 & 3\\ 1 & -1 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 2\ +\ 3) -\overrightarrow{j}(4\ -\ 3)+\overrightarrow{k}(-4\ -\ 2)$
$= \overrightarrow{i}(5) -\overrightarrow{j}(1)\ +\ \overrightarrow{k}(- 6)$
$\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}\ =\ 5\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 6\overrightarrow{k}$
$Area\ of \ parellelogram =\ \frac{1}{2} |\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}|$
$=\ \frac{1}{2} \sqrt{(5)^2 + (-1)^2 + (-6)^2 }$
$=\ \frac{1}{2}\ \sqrt{(25\ +\ 1\ +\ 36)}$
$=\ \frac{1}{2}\ \sqrt{62}$
$\boxed{Area\ of \ parellelogram =\ \frac{1}{2}\ \sqrt{62} sq.units}$

$14.\ Find\ the\ area\ of\ the\ triangle\ whose\ adjacent\ sides\ are\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k} and\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}.$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}=2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 2 & 3 & -1\\ 1 & 3 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}(3\ +\ 3)\ -\ \overrightarrow{j}(2\ +\ 1)\ +\ \overrightarrow{k}(6\ -\ 3)$
$= \overrightarrow{i}(6)\ -\ \overrightarrow{j}(3)\ +\ \overrightarrow{k}(3)$
$\overrightarrow{a}× \overrightarrow{b}\ =\ 6\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$Area\ of \ triangle\ =\ \frac{1}{2} |\overrightarrow{a}\ ×\ \overrightarrow{b}|$
$=\ \frac{1}{2} \sqrt{(6)^2 + (-3)^2 + (3)^2 }$
$=\ \frac{1}{2}\ \sqrt{(36\ +\ 9\ +\ 9)}$
$=\ \frac{1}{2}\ \sqrt{54}$

$\underline{PART\ -\ C}$
$15.\ If\ \overrightarrow{a}\ = \ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ \overrightarrow{i} -\ 4\overrightarrow{j}\ -\ 6\overrightarrow{k},\ \hspace{15cm}$$find\ the\ projection\ of\ \overrightarrow{a}\ on\ \overrightarrow{b}\ .\ Also\ find\ the\ angle\ between\ them\ \hspace{5cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{b}\ =\ \overrightarrow{i} -\ 4\overrightarrow{j}\ -\ 6\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= ( 2\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 3\overrightarrow{k}) .(\overrightarrow{i} -\ 4\overrightarrow{j}\ -\ 6\overrightarrow{k})$

= 2 ( 1 ) + 1 ( -4) + 3 ( – 6 )

=    2 – 4 – 18

=   – 20

$\boxed{ \overrightarrow{a}.\overrightarrow{b}\ =\ -\ 20}$
$\overrightarrow{|a|} = \sqrt{(2)^2 + (1)^2 + (3)^2 }=\sqrt{4\ +\ 1\ +\ 9 }=\sqrt{14}$
$\overrightarrow{|b|} = \sqrt{(1)^2 + (-4)^2 + (-6)^2 }=\sqrt{1\ +\ 16\ +\ 36 }=\sqrt{53}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} = \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}} = \frac{-20}{ \sqrt{53}}$
$\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b}\ =\ \frac{-20}{ \sqrt{53}}}$
$\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$= \frac{-20}{\sqrt{14}\sqrt{53}}$
$\boxed{\theta = \cos ^{-1} ( \frac{-20}{\sqrt{14}\sqrt{53}})}$

$16.\ Prove\ that\ the\ vectors\ \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}\ and\ 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}\ \hspace{15cm}$$are\ mutually\ perpendicular.\ \hspace{5cm}$

Soln:

$\overrightarrow{a}= \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{c}= 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}) .(\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k})$

= 1(1) + 2(1) + 1(-3)

= 0

$\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.$
$\overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}) .( 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})$

= 1(7) + 1(-4) – 3(1)

= 7 – 4 – 3

= 0

$\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.$

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

$\overrightarrow{c}.\overrightarrow{a}= (7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}) .(\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k})$

= 7 – 8 + 1

= 0

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

$17.\ Show\ that\ (\overrightarrow{a}\ .\ \overrightarrow{i})\overrightarrow{i}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{j})\overrightarrow{j}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{k})\overrightarrow{k}\ =\ \overrightarrow{a}\ ,\ if\ \overrightarrow{a}\ is\ any\ vector\ \hspace{10cm}$
$Soln:\ \hspace{20cm}$
$Let\ \overrightarrow{a}\ =\ x\overrightarrow{i}\ +\ y\overrightarrow{j}\ +\ z\overrightarrow{k}$
$\overrightarrow{a}\ .\ \overrightarrow{i}\ =\ (x\overrightarrow{i}\ +\ y\overrightarrow{j}\ +\ z\overrightarrow{k})\ .\ \overrightarrow{i}$
$=\ x$
$\overrightarrow{a}\ .\ \overrightarrow{j}\ =\ (x\overrightarrow{i}\ +\ y\overrightarrow{j}\ +\ z\overrightarrow{k})\ .\ \overrightarrow{j}$
$=\ y$
$\overrightarrow{a}\ .\ \overrightarrow{k}\ =\ (x\overrightarrow{i}\ +\ y\overrightarrow{j}\ +\ z\overrightarrow{k})\ .\ \overrightarrow{k}$
$=\ z$
$L.\ H.\ S\ =\ (\overrightarrow{a}\ .\ \overrightarrow{i})\overrightarrow{i}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{j})\overrightarrow{j}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{k})\overrightarrow{k}$
$=\ x\overrightarrow{i}\ +\ y\overrightarrow{j}\ +\ z\overrightarrow{k}$
$=\ \overrightarrow{a}\ =\ R.\ H.\ S$
$\boxed{(\overrightarrow{a}\ .\ \overrightarrow{i})\overrightarrow{i}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{j})\overrightarrow{j}\ +\ (\overrightarrow{a}\ .\ \overrightarrow{k})\overrightarrow{k}\ =\ \overrightarrow{a}}$

$18\ Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ \hspace{15cm}$$5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k}\ ,\ \overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k} and\ -\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}\ \hspace{5cm}$

Soln:

$\overrightarrow{OA}= 5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k}$
$\overrightarrow{OB}= \overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{OC}=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}- (5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k})$
$=\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}- 5\overrightarrow{i}-2\overrightarrow{j} – 4\overrightarrow{k})$
$\overrightarrow{AB}= -4\overrightarrow{i} + \overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}- (\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k})$
$=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}- \overrightarrow{i} -3\overrightarrow{j} – 2\overrightarrow{k})$
$\overrightarrow{BC}= -2\overrightarrow{i} – 4\overrightarrow{j} – \overrightarrow{k}$
$\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -4 & 1 & -2\\ -2 & -4 & -1\\ \end{vmatrix}$
$= \overrightarrow{i}( -1 – 8) -\overrightarrow{j}(4 – 4)+\overrightarrow{k}(16 + 2)$
$= \overrightarrow{i}(-9) – 0\overrightarrow{j}(-8)+\overrightarrow{k}(18)$
$\overrightarrow{AB}× \overrightarrow{BC}= -9\overrightarrow{i}+18\overrightarrow{k}$
$|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(9 )^2 + (18)^2}=\sqrt{(81 + 324 }=\sqrt{405}$
$Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|$
$=\frac{\sqrt{405}}{2}\ sq. units$

$19.\ \color{red}{Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors}\ \hspace{15cm}$$\color{red}{3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\ ,\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ and\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}}\ \hspace{5cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{OB}\ =\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}$
$\overrightarrow{OC}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k})$
$=\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ – 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}$
$\boxed{\overrightarrow{AB}\ = -2\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}\ – (\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k})$
$=\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}\ – \overrightarrow{i}\ +\ 3\overrightarrow{j}\ -5\overrightarrow{k}$
$\boxed{\overrightarrow{BC}\ =\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ 9\overrightarrow{k}}$
$\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -2 & -1 & 4\\ 1 & 4 & – 9\\ \end{vmatrix}$
$= \overrightarrow{i}(9\ -\ 16)\ -\overrightarrow{j}(18\ -\ 4)\ +\ \overrightarrow{k}(-\ 8 +\ 1)$
$= \overrightarrow{i}(-7)\ -\overrightarrow{j}(14)\ +\ \overrightarrow{k}(-\ 7)$
$\boxed{\overrightarrow{AB}× \overrightarrow{BC}\ =\ -\ 7\overrightarrow{i}\ -\ 14\overrightarrow{j}\ -\ 7\overrightarrow{k}}$
$|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-7 )^2\ +\ (-14)^2\ +\ ((-7)^2 }=\sqrt{(49\ +\ 196\ +\ 49}=\sqrt{294}$
$Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|$
$\boxed{Area\ of \ triangle\ =\frac{\sqrt{294}}{2}\ sq. units}$

$20.\ Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ \hspace{15cm}$$3\overrightarrow{i}\ +\ 3\overrightarrow{j}+ \overrightarrow{k}\ and\ 2\overrightarrow{i}\ -\ 5\overrightarrow{j}\ +\ 3\overrightarrow{k}.\ \hspace{5cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}\ +\ 3\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i}\ -\ 5\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 3 & 3 & 1\\ 2 & – 5 & 3\\ \end{vmatrix}$
$= \overrightarrow{i}( 9\ +\ 5)\ -\ \overrightarrow{j}(9\ -\ 2)\ +\ \overrightarrow{k}(-\ 15\ -\ 6)$
$= \overrightarrow{i}(14)\ -\ \overrightarrow{j}(7)\ +\ \overrightarrow{k}(-21)$
$\overrightarrow{a}× \overrightarrow{b}\ =\ 14\overrightarrow{i}\ -\ 7\overrightarrow{j}\ -\ 21\overrightarrow{k}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(14 )^2 + (-7)^2 + (-21)^2 }=\sqrt{(196\ +\ 49\ +\ 441}\ =\ \sqrt{686}$
$n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}\ =\ \frac{14\overrightarrow{i} -\ 7\overrightarrow{j}\ -\ 21\overrightarrow{k}}{\sqrt{686}}$
$\boxed{n^\wedge = \frac{14\overrightarrow{i} -\ 7\overrightarrow{j}\ -\ 21\overrightarrow{k}}{\sqrt{686}}}$

$21.\ Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ \overrightarrow{i} – \overrightarrow{j}+ 3\overrightarrow{k} and\ 2\overrightarrow{i}+ 3\overrightarrow{j} -\overrightarrow{k}.\ \hspace{10cm}$$Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\ \hspace{10cm}$

Soln:

$\overrightarrow{a}= \overrightarrow{i} – \overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{b}= 2\overrightarrow{i}+ 3\overrightarrow{j} -\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & -1 & 3\\ 2 & 3 & -1\\ \end{vmatrix}$
$= \overrightarrow{i}( 1 – 9) -\overrightarrow{j}(-1 – 6)+\overrightarrow{k}(3 + 2)$
$= \overrightarrow{i}(-8) -\overrightarrow{j}(-7)+\overrightarrow{k}(5)$
$\overrightarrow{a}× \overrightarrow{b}= -8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-8 )^2 + (-7)^2 + (5)^2 }=\sqrt{(64 + 49 + 25}=\sqrt{138}$
$n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{-8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}}{\sqrt{138}}$
$n^\wedge = \frac{-8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}}{\sqrt{138}}$
$\overrightarrow{|a|} = \sqrt{(1)^2 + (-1)^2 + (3)^2 }=\sqrt{(1 + 1 + 9 }=\sqrt{11}$
$\overrightarrow{|b|} = \sqrt{(2)^2 + (3)^2 + (-1)^2 }=\sqrt{(4 + 9 + 1 }=\sqrt{14}$
$\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{138}}{\sqrt{11}\sqrt{14}}$
$\ sin\ \theta =\frac{\sqrt{138}}{\sqrt{11}\sqrt{14}}$
$22.\ The\ forces\ 2\overrightarrow{i}\ -\ 5\overrightarrow{j}\ + \ 6\overrightarrow{k}\ ,\ -\overrightarrow{i}\ +\ 2\overrightarrow{j}\ – \ \overrightarrow{k}\ and\ 2\overrightarrow{i}\ +\ 7\overrightarrow{j}\ act\ on\ a\ particle\ \hspace{15cm}$$and\ displace\ it\ from\ the\ point\ 4\overrightarrow{i}\ -\ 3\overrightarrow{j}\ -\ 2\overrightarrow{k}\ to\ the\ point\ 6\overrightarrow{i}\ +\ \overrightarrow{j}\ – \ 3\overrightarrow{k}.\ Find\ the\ total\ work\ done\ by\ the\ forces\ \hspace{8cm}$
$Soln:\ \hspace{18cm}$
$\overrightarrow{F_1}= 2\overrightarrow{i}\ -\ 5\overrightarrow{j}\ + \ 6\overrightarrow{k}$
$\overrightarrow{F_2}= -\overrightarrow{i}\ +\ 2\overrightarrow{j}\ – \ \overrightarrow{k}$
$\overrightarrow{F_3}= 2\overrightarrow{i}\ +\ 7\overrightarrow{j}$
$\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2}\ +\ \overrightarrow{F_3}\ =\ 3\overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k} + \overrightarrow{i}+ 7\overrightarrow{j}+7\overrightarrow{k}\ +\ 2\overrightarrow{i}\ +\ 7\overrightarrow{j}$
$\boxed{\overrightarrow{F}= 3\overrightarrow{i}\ + 4\overrightarrow{j}\ +\ 5\overrightarrow{k}}$
$\overrightarrow{OA}= 4\overrightarrow{i}\ -\ 3\overrightarrow{j}\ -\ 2\overrightarrow{k}$
$\overrightarrow{OB}= 6\overrightarrow{i}\ +\ \overrightarrow{j}\ – \ 3\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=6\overrightarrow{i}\ +\ \overrightarrow{j}\ – \ 3\overrightarrow{k}- (4\overrightarrow{i}\ -\ 3\overrightarrow{j}\ -\ 2\overrightarrow{k})$
$=6\overrightarrow{i}\ +\ \overrightarrow{j}\ – \ 3\overrightarrow{k}\ – 4\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\boxed{\overrightarrow{d}\ =\ 2\overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ \overrightarrow{k}}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}= (3\overrightarrow{i}\ + 4\overrightarrow{j}\ +\ 5\overrightarrow{k}) .(2\overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ \overrightarrow{k})$
$=\ 3 ( 2 )\ +\ 4 ( 4 )\ + 5 ( -1 )$
$=\ 6\ +\ 16\ -\ 5$
$\boxed{Work\ done\ =\ 17\ units}$
$23.\ The\ forces\ 3\overrightarrow{i}\ +\ 5\overrightarrow{j}\ – \ 2\overrightarrow{k}\ and\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ 5\overrightarrow{k}\ displaces\ a\ particle\ \hspace{15cm}$$from\ the\ point\ (1, 2, -1)\ to\ the\ point\ (5, -3, 4).\ Find\ the\ total\ work\ done\ by\ the\ force.\ \hspace{8cm}$
$Soln:\ \hspace{18cm}$
$\overrightarrow{F_1}= 3\overrightarrow{i}\ +\ 5\overrightarrow{j}\ – \ 2\overrightarrow{k}$
$\overrightarrow{F_2}= 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ 5\overrightarrow{k}$
$\overrightarrow{F}=\overrightarrow{F_1} + \overrightarrow{F_2} = 3\overrightarrow{i}\ +\ 5\overrightarrow{j}\ – \ 2\overrightarrow{k} + 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ 5\overrightarrow{k}$
$\boxed{\overrightarrow{F}\ =\ 5\overrightarrow{i}\ + 8\overrightarrow{j}\ -\ 7\overrightarrow{k}}$
$\overrightarrow{OA}= \overrightarrow{i}+ 2\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OB}\ =\ 5\overrightarrow{i}\ -3\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=\ 5\overrightarrow{i}\ -3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ – (\overrightarrow{i}+ 2\overrightarrow{j}\ -\ \overrightarrow{k})$
$=5\overrightarrow{i}\ -3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ – \overrightarrow{i}\ – 2\overrightarrow{j}\ + \overrightarrow{k}$
$\boxed{\overrightarrow{d}\ =\ 4\overrightarrow{i}\ – 5\overrightarrow{j}\ +\ 5\overrightarrow{k}}$
$Work\ done = \overrightarrow{F}.\overrightarrow{d}\ = (5\overrightarrow{i}\ + 8\overrightarrow{j}\ -\ 7\overrightarrow{k}) .(4\overrightarrow{i}\ – 5 \overrightarrow{j}\ +\ 5\overrightarrow{k})$
$=\ 5( 4 )\ +\ 8 ( -5 )\ -7 ( 5 )$
$=\ 20\ -\ 40\ -\ 35$
$Work\ done\ =\ -\ 55$
$\boxed{Work\ done\ =\ 55\ units}\ (by\ taking\ positive\ value)$
$24.\ \color{red}{Find\ the\ moment\ of\ the\ force\ 3\overrightarrow{i}+4\overrightarrow{j}+5\overrightarrow{k}\ acting\ through\ the\ point}\ \hspace{10cm}$$\color{red}{\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k} about\ the\ point\ 4\overrightarrow{i}- 3\overrightarrow{j}+\overrightarrow{k}.}\ \hspace{10cm}$
$Soln:\ \hspace{18cm}$
$\overrightarrow{F}= 3\overrightarrow{i} +4\overrightarrow{j} + 5\overrightarrow{k}$
$\overrightarrow{OP}= \overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{OA}= 4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k}$
$\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}$
$=\overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}- (4\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})$
$=\overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}- 4\overrightarrow{i} +3\overrightarrow{j} -\overrightarrow{k}$
$\overrightarrow{r}= -3\overrightarrow{i} + \overrightarrow{j} + 2\overrightarrow{k}$
$Moment = \overrightarrow{r}× \overrightarrow{F}$
$=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -3 & 1 & 2\\ 3 & 4 & 5\\ \end{vmatrix}$
$= \overrightarrow{i}( 5 – 8) -\overrightarrow{j}(-15 – 6)+\overrightarrow{k}(-12 – 3)$
$\overrightarrow{r}× \overrightarrow{F}= -3\overrightarrow{i}+ 21\overrightarrow{j}-15\overrightarrow{k}$
$Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|$
$= \sqrt{(-3)^2 + (21)^2 + (-15)^2 }=\sqrt{(9 + 441+ 225 }=\sqrt{675}$
$\boxed{Magnitude\ of \ Moment = \sqrt{675}}$