# SOLUTIONS TO SEMINAR – 1 FOR ENGINEERING MATHEMATICS – II

$1.\ Find\ the\ equation\ of\ the\ circle\ passing\ through\ the\ point\ A(2.3)\ \hspace{7cm}$$and\ having\ its\ centre\ at\ (4,1)\ \hspace{5cm}$
$Soln:\ r\ =\ \sqrt{(4\ -\ 2)^2\ +\ (1\ -\ 3)^2}\ \hspace{15cm}$
$=\ \sqrt{(2)^2\ +\ (-\ 2)^2}\ \hspace{10cm}$
$=\ \sqrt{4\ +\ 4}\ \hspace{10cm}$
$r=\ \sqrt{8}\ \hspace{10cm}$
$We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -k)^2\ =\ r^2\ \hspace{7cm}$
$Here\ h\ =\ 4,\ k\ =\ 1\ (Given)\ and\ r\ =\ \sqrt{8}\ \hspace{10cm}$
$(x\ -\ 4)^2\ +\ (y\ -\ 1)^2\ =\ (\sqrt{8})^2\ \hspace{10cm}$
$x^2\ -\ 8\ x\ +\ 16\ +\ y^2\ -\ 2y\ +\ 1\ =\ 8\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 8\ x\ -\ 2y\ +\ 17\ -\ 8\ =\ 0\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 8\ x\ -\ 2y\ +\ 9\ =\ 0\ \hspace{10cm}$
$\therefore\ the\ required\ equation\ of\ the\ circle\ is\ \hspace{7cm}$
$\boxed{x^2\ +\ y^2\ -\ 8\ x\ -\ 2\ y\ +\ 9\ =\ 0}\ \hspace{5cm}$
$2.\ Prove\ that\ the\ circles\ x^2\ +\ y^2\ -\ 8\ x\ +\ 6y\ -\ 23\ =\ 0\ and\ \hspace{7cm}$$and\ x^2\ +\ y^2\ -\ 2\ x\ -\ 5y\ +\ 16\ =\ 0\ cut\ orthogonally\ \hspace{5cm}$
$Soln:\ Given\ x^2\ +\ y^2\ -\ 8\ x\ +\ 6y\ -\ 23\ =\ 0\ —–(1)\ and\ \hspace{10cm}$
$x^2\ +\ y^2\ -\ 2\ x\ -\ 5y\ +\ 16\ =\ 0\ —–(2)\ \hspace{5cm}$
$From\ (1)\ \hspace 10cm$
$2g_1 =\ -\ 8\ \hspace 2cm\ 2f_1 = 6\ \hspace 2cm\ c_1 = – 23$
$g_1 =\ -\ 4\ \hspace 2cm\ f_1 =\ 3\ \hspace 2cm$
$From\ (2)\ \hspace 10cm$
$2g_2 = -\ 2\ \hspace 2cm\ 2f_2 =\ -\ 5\ \hspace 2cm\ c_2 =\ 16$
$g_2 = \ 1\ \hspace 2cm\ f_2 =\ -\ \frac{5}{2}\ \hspace 2cm$
$The\ condition\ for\ orthogonally\ is\ \hspace{10cm}$
$2\ g_1\ g_2\ +\ 2\ f_1\ f_2\ =\ c_1\ +\ c_2\ \hspace{10cm}$
$2\ (-4)\ (-1)\ +\ 2\ (3)\ (-\ \frac{5}{2})\ =\ -\ 23\ +\ 16\ \hspace{10cm}$
$8\ -\ 15\ =\ -\ 7\ \hspace{10cm}$
$-\ 7\ =\ -\ 7\ \hspace{10cm}$
$\therefore\ The\ circles\ cut\ orthogonally$
$3.\ Prove\ that\ equation\ 2\ x^2\ -\ 7\ x\ y\ +\ 3\ y^2\ +\ 5\ x\ -\ 5\ y\ +\ 2\ =\ 0\ \hspace{7cm}$$represents\ a\ pair\ of\ straight\ lines\ \hspace{5cm}$
$Soln:\ Given\ 2\ x^2\ -\ 7\ x\ y\ +\ 3\ y^2\ +\ 5\ x\ -\ 5\ y\ +\ 2\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}$
$\hspace{2cm}\ a\ =\ 2\ \hspace{2cm}\ 2\ h\ =\ – 7\ \hspace{2cm}\ b\ =\ 3\ \hspace{2cm}\ 2\ g\ =\ 5\ \hspace{2cm}\ 2\ f\ =\ -\ 5\ \hspace{2cm}\ c\ =\ 2$
$\hspace{6cm}\ h\ =\ \frac{-7}{2}\ \hspace{4cm}\ g\ =\ \frac{5}{2}\ \hspace{4cm}\ f\ =\ \frac{-5}{2}\ \hspace{4cm}$
$\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 2\ (3)\ (2)\ +\ 2\ (\frac{-5}{2})\ (\frac{5}{2})\ (\frac{-7}{2})\ -\ 2\ (\frac{-5}{2}) ^2\ -\ 3\ (\frac{5}{2}) ^2\ -\ 2\ (\frac{-7}{2}) ^2\ \hspace{8cm}$
$\hspace{2cm}\ =\ 12\ +\ \frac{175}{4}\ -\ 2\ (\frac{25}{4})\ -\ 3\ (\frac{25}{4})\ -\ 2\ (\frac{49}{4})\ \hspace{10cm}$
$=\ 12\ +\ \frac{175}{4}\ -\ \frac{50}{4}\ -\ \frac{75}{4}\ -\ \frac{98}{4}\ \hspace{10cm}$
$=\ 12\ +\ (\frac{175\ -\ 50\ -\ 75\ -\ 98}{4})\ \hspace{10cm}$
$=\ 12\ +\ (\frac{175\ -\ 223}{4})\ \hspace{10cm}$
$=\ 12\ +\ \frac{- 48}{4}\ \hspace{10cm}$
$=\ 12\ -\ 12\ \hspace{10cm}$
$=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}$
$\hspace{2cm}\ \therefore\ The\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}$
$4.\ Show\ that\ the\ points\ whose\ position\ vectors\ \hspace{15cm}$$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\ \hspace{5cm}$
$Soln:\ Given\ \hspace{20cm}$
$\overrightarrow{OA}= 2\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{OC}= -\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}- (2\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k})$
$=3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}- 2\overrightarrow{i}\ + \overrightarrow{j} – 3\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} – 4\overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=-\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}- (3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k})$
$=-\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}- 3\overrightarrow{i}\ + 5\overrightarrow{j} – \overrightarrow{k}$
$\overrightarrow{BC}= -4\overrightarrow{i} + 16\overrightarrow{j} + 8\overrightarrow{k}$
$\overrightarrow{BC}= -4(\overrightarrow{i} – 4\overrightarrow{j} – 2\overrightarrow{k})$
$\overrightarrow{BC} = -4\overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$
$5.\ Prove\ that\ the\ points\ \hspace{15cm}$
$2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k}, 3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k} and\ 4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}\ form\ an\ equilateral\ triangle$
$Soln:\ Given\ \hspace{20cm}$
$\overrightarrow{OA}= 2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k}$
$\overrightarrow{OC}= 4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k}- (2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k})$
$=3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k}- 2\overrightarrow{i}\ – 3\overrightarrow{j}- 4\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} +\overrightarrow{j} – 2\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }$
$= \sqrt{(1 + 1 +4) }$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}- (3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k})$
$=4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}- 3\overrightarrow{i}\ – 4\overrightarrow{j}- 2\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} – 2\overrightarrow{j} + \overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (- 2)^2 +(1)^2 }$
$= \sqrt{(1 + 4 + 1)}$
$BC = \sqrt{6}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}- (2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k})$
$=4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}- 2\overrightarrow{i}\ – 3\overrightarrow{j}- 4\overrightarrow{k}$
$\overrightarrow{AC}= 2\overrightarrow{i} -\overrightarrow{j} -\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(2)^2 + (-1)^2 +(-1)^2 }$
$= \sqrt{(4 + 1 + 1)}$
$AC = \sqrt{6}$
$AB = BC = AC = \sqrt{6}$
$\therefore\ The\ given\ points\ represent\ equilateral\ triangle$
$6.\ If\ \overrightarrow{a}\ = \ 8\overrightarrow{i}\ +\ 4\overrightarrow{j}\ – 3\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k},\ \hspace{15cm}$$find\ the\ projection\ of\ \overrightarrow{a}\ on\ \overrightarrow{b}\ .\ Also\ find\ the\ angle\ between\ them\ \hspace{5cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}= 8\overrightarrow{i}\ +\ 4\overrightarrow{j}\ – 3\overrightarrow{k}$
$\overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{a}.\overrightarrow{b}= (8\overrightarrow{i}+ 4\overrightarrow{j}- 3\overrightarrow{k}) .(2\overrightarrow{i} – 3 \overrightarrow{j} +2\overrightarrow{k})$
$=\ 8(2)\ +\ 4(-\ 3)\ -\ 3(2)$
$=\ 16\ -\ 12\ -\ 6$
$=\ -\ 2$
$\boxed{ \overrightarrow{a}.\overrightarrow{b}\ =\ -\ 2}$
$\overrightarrow{|a|} = \sqrt{(8)^2 + (4)^2 + (-3)^2 }=\sqrt{64\ +\ 16\ +\ 9 }=\sqrt{89}$
$\overrightarrow{|b|} = \sqrt{(2)^2 + (-3)^2 + (2)^2 }=\sqrt{4\ +\ 9\ +\ 4 }=\sqrt{17}$
$Projection\ of\ \overrightarrow{a} on \overrightarrow{b} = \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}} = \frac{-2}{ \sqrt{17}}$
$\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b}\ =\ \frac{-2}{ \sqrt{17}}}$
$\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}$
$= \frac{-2}{\sqrt{89}\sqrt{17}}$
$\boxed{\theta = \cos ^{-1} ( \frac{-2}{\sqrt{89}\sqrt{17}})}$
$7.\ Find\ the\ work\ done\ by\ the\ force\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ \overrightarrow{k}\ when\ it\ displaces\ a\ particle\ \hspace{15cm}$$from\ the\ point\ 2\overrightarrow{i}\ -\ 6\overrightarrow{j}\ +\ 7\overrightarrow{k}\ to\ the\ point\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ – \ 5\overrightarrow{k}\ \hspace{8cm}$
$Soln:\ \hspace{18cm}$
$\overrightarrow{F}\ =\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ \overrightarrow{k}$
$\overrightarrow{OA}= 2\overrightarrow{i}\ -\ 6\overrightarrow{j}\ +\ 7\overrightarrow{k}$
$\overrightarrow{OB}\ = 3\overrightarrow{i}\ -\ \overrightarrow{j}\ – \ 5\overrightarrow{k}$
$\overrightarrow {d}= \overrightarrow {OB}- \overrightarrow{OA}$
$=\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ – \ 5\overrightarrow{k}\ – (2\overrightarrow{i}\ -\ 6\overrightarrow{j}\ +\ 7\overrightarrow{k})$
$=3\overrightarrow{i}\ -\ \overrightarrow{j}\ – \ 5\overrightarrow{k}\ – 2\overrightarrow{i}\ + 6\overrightarrow{j}\ – 7\overrightarrow{k}$
$\boxed{\overrightarrow{d}\ =\ \overrightarrow{i}\ +\ 5\overrightarrow{j}\ -\ 12\overrightarrow{k}}$
$Work\ done\ = \overrightarrow{F}.\overrightarrow{d}= (\overrightarrow{i}\ +\ 3\overrightarrow{j}\ – \ \overrightarrow{k}) .(\overrightarrow{i}\ +\ 5\overrightarrow{j}\ -\ 12\overrightarrow{k})$
$=\ 1 ( 1 )\ +\ 3 ( 5 )\ – 1 ( -12 )$
$=\ 1\ +\ 15\ +\ 12$
$\boxed{Work\ done\ =\ 28\ units}$
$8.\ Find\ the\ moment\ of\ the\ force\ 6\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ \overrightarrow{k}\ acting\ through\ the\ point\ \hspace{10cm}$$\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k} about\ the\ point\ -\overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k}.\ \hspace{10cm}$
$Soln:\ \hspace{18cm}$
$\overrightarrow{F}= 6\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{OP}= \overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{OA}\ =\ -\overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}$
$=\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – (-\overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k})$
$=\overrightarrow{i} + 2\overrightarrow{j} + 3\overrightarrow{k}\ +\ \overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\boxed{\overrightarrow{r}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}}$
$Moment = \overrightarrow{r}× \overrightarrow{F}$
$=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 2 & 3 & 2\\ 6 & 1 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 3 – 2) -\overrightarrow{j}(2 – 12)+\overrightarrow{k}(2 – 18)$
$\boxed{\overrightarrow{r}× \overrightarrow{F}=\ \overrightarrow{i}\ +\ 10\overrightarrow{j}\ -\ 16\overrightarrow{k}}$
$Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|$
$= \sqrt{(1)^2 + (10)^2 + (-16)^2 }=\sqrt{(1\ +\ 100\ +\ 256 }=\sqrt{357}$
$\boxed{Magnitude\ of \ Moment = \sqrt{357}\ units}$