# SOLUTIONS TO ASSIGNMENT – I FOR ENGINEERING MATHEMATICS – II

$1.\ Show\ that\ the\ circles\ x^2 + y^2 – 4x + 6y – 8 = 0\ and\ \hspace{10cm}$$x^2 + y^2 – 10x – 6y +\ 14\ =\ 0\ touch\ each\ other.\ \hspace{10cm}$

Soln:    Given  x2  +   y2   – 4x  +  6y  + 8 = 0 ––––– (1)      and

x2  +   y– 10x  –  6y  +14 = 0  ––––– (2)

From ( 1 )

2g1  = -4                    2f1 = 6               c1 = 8

g1 =  -2                       f1= 3

centre  is  C1  = ( – g1 , – f1 )                 r1 = √( g12  + f12 – c)

=   ( 2 , -3 )                 r= √( (-2)2  + (3)2 – 8)

r= √( 4  + 9  – 8 )

r1 = √13 – 8 = √5

From ( 2  )

2g2   = -10                    2f2 = -6                c2 = 14

g2 =  – 5                          f2= – 3

centre  is  C2  = ( – g2 , – f2 )                 r2 = √( g22  + f22 – c)

=   ( 5  , 3 )                  r= √( (-5)2  + (-3)2  – 14)

r= √( 25  + 9  – 14 )

r2 = √20 =   √(4 × 5)  =   2√5

C2=   ( 5   , 3 )       &   r2 =   2√5

C1C2   =  √(( 5 – 2 )2 + (3 + 3 )2)

=   √( (3)2  + (6)2 )

= √( 9  + 36)

= √45 =   √(9 × 5)   =   3√5

C1C2   =   3√5

r1  +  r2  =   √5   +   2√5

=   3√5   =   C1C2

∴The circles touch each other externally.

$2.\ Find\ ‘α’ \ such\ that\ the\ equation\ \ 3\ x^2\ +\ 7\ x\ y\ +\ α\ y^2\ -\ 4\ x\ -\ 13\ y\ -\ 7\ =\ 0\ \hspace{7cm}$$represents\ a\ pair\ of\ straight\ lines\ \hspace{5cm}$
$Soln:\ Given\ 3\ x^2\ +\ 7\ x\ y\ +\ α\ y^2\ -\ 4\ x\ -\ 13\ y\ -\ 7\ =\ 0\ ———-(1)\ \hspace{8cm}$
$\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}$
$comparing\ we\ get\ \hspace{10cm}$
$\hspace{2cm}\ a\ =\ 3\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ α\ \hspace{2cm}\ 2\ g\ =\ -\ 4\ \hspace{2cm}\ 2\ f\ =\ -\ 13\ \hspace{2cm}\ c\ =\ -\ 7$
$\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{6cm}\ g\ =\ -\ 2\ \hspace{4cm}\ f\ =\ \frac{- 13}{2}\ \hspace{4cm}$
$\hspace{2cm}\ Given\ that\ \ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}$
$\hspace{2cm}\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}$
$\hspace{2cm}\ -\ 21\ α\ +\ 91\ -\ 3\ (\frac{169}{4})\ -\ 4\ (α)\ +\ 7\ (\frac{49}{4})\ =\ 0\ \hspace{10cm}$
$\hspace{2cm}\ \frac{-\ 84 α\ +\ 364\ -\ 507\ -\ 16\ α\ +\ 343}{4}\ =\ 0\ \hspace{10cm}$
$100\ α\ +\ 707\ -\ 507=\ 0\ \hspace{10cm}$
$100\ α\ +\ 200\ =\ 0\ \hspace{10cm}$
$100\ α\ =\ -\ 200\ \hspace{10cm}$
$\boxed {α\ =\ 2}\ \hspace{10cm}$
$3.\ The\ position\ vectors\ of\ the\ \triangle\ ABC\ are\ \hspace{18cm}$$2\overrightarrow{i}\ – \overrightarrow{j}+ \overrightarrow{k}, \overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k} and\ 3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}\ respectively.\ Prove\ that\ triangle\ is\ right\ angled\ \hspace{10cm}$

Soln: Given

$\overrightarrow{OA}= 2\overrightarrow{i}\ – \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{OB}= \overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k}$
$\overrightarrow{OC}= 3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k}- (2\overrightarrow{i}\ – \overrightarrow{j}+ \overrightarrow{k})$
$=\overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k}- 2\overrightarrow{i}\ + \overrightarrow{j} – \overrightarrow{k}$
$\overrightarrow{AB} = -\overrightarrow{i} – 2\overrightarrow{j} – 6\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(-1)^2 + (-2)^2 +(-6)^2 }$
$= \sqrt{(1 + 4 +36 }$
$AB = \sqrt{41}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}- (\overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k})$
$=3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}- \overrightarrow{i}\ + 3\overrightarrow{j} + 5\overrightarrow{k})$
$\overrightarrow{BC}= 2\overrightarrow{i} -\overrightarrow{j} + \overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(2)^2 + (-1)^2 +(1)^2 }$
$= \sqrt{(4 + 1 + 1}$
$BC = \sqrt{6}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}- (2\overrightarrow{i}\ – \overrightarrow{j}+ \overrightarrow{k})$
$=3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}- 2\overrightarrow{i}\ + \overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{AC}= \overrightarrow{i} -3\overrightarrow{j} -5\overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(1)^2 + (-3)^2 +(-5)^2 }$
$= \sqrt{(1 + 9+ 25}$
$AC = \sqrt{35}$
$AB^2 = 41, BC^2 = 6, AC^2 = 35$
$BC^2 + AC^2 = 6 + 35 = 41=AB^2$
$BC^2 + AC^2 = AB^2$

The given triangle is an right angled triangle.

$4.\ Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ 3\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k} and\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}.\ \hspace{10cm}$$Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\ \hspace{10cm}$
$Soln:\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{b}=4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 3 & -3 & 2\\ 4 & -2 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}(-3 + 4)\ -\overrightarrow{j}(3 – 8)\ +\ \overrightarrow{k}(-6 + 12)$
$= \overrightarrow{i}(1)\ -\ \overrightarrow{j}(-5)\ +\ \overrightarrow{k}(6)$
$\boxed{\overrightarrow{a}× \overrightarrow{b}\ =\ \overrightarrow{i}\ +\ 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}$
$|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(1 )^2 + (5)^2 + (6)^2 }=\sqrt{1\ +\ 25\ +\ 36}=\sqrt{62}$
$n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{\overrightarrow{i}\ + 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}{\sqrt{62}}$
$\boxed{n^\wedge = \frac{\overrightarrow{i}\ + 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}{\sqrt{62}}}$
$\overrightarrow{|a|} = \sqrt{(3)^2 + (-3)^2 + (2)^2 }=\sqrt{(9\ +\ 9\ +\ 4}=\sqrt{22}$
$\overrightarrow{|b|} = \sqrt{(4)^2 + (-2)^2 + (1)^2 }=\sqrt{(16\ +\ 4\ +\ 1}=\sqrt{21}$
$\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{62}}{\sqrt{22}\sqrt{21}}$
$\boxed{sin\ \theta =\frac{\sqrt{62}}{\sqrt{22}\sqrt{21}}}$