\[1.\ Show\ that\ the\ circles\ x^2 + y^2 – 4x + 6y – 8 = 0\ and\ \hspace{10cm}\]\[x^2 + y^2 – 10x – 6y +\ 14\ =\ 0\ touch\ each\ other.\ \hspace{10cm}\]
Soln: Given x2 + y2 – 4x + 6y + 8 = 0 ––––– (1) and
x2 + y2 – 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 ) r1 = √( (-2)2 + (3)2 – 8)
r1 = √( 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 ) r2 = √( (-5)2 + (-3)2 – 14)
r2 = √( 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.
\[4.\ Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ 3\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k} and\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}.\ \hspace{10cm}\]\[Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\ \hspace{10cm}\]
\[Soln:\ \hspace{20cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k} \]
\[\overrightarrow{b}=4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
3 & -3 & 2\\
4 & -2 & 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}(-3 + 4)\ -\overrightarrow{j}(3 – 8)\ +\ \overrightarrow{k}(-6 + 12)\]
\[ = \overrightarrow{i}(1)\ -\ \overrightarrow{j}(-5)\ +\ \overrightarrow{k}(6)\]
\[\boxed{\overrightarrow{a}× \overrightarrow{b}\ =\ \overrightarrow{i}\ +\ 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(1 )^2 + (5)^2 + (6)^2 }=\sqrt{1\ +\ 25\ +\ 36}=\sqrt{62}\]
\[ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{\overrightarrow{i}\ + 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}{\sqrt{62}} \]
\[\boxed{n^\wedge = \frac{\overrightarrow{i}\ + 5\overrightarrow{j}\ +\ 6\overrightarrow{k}}{\sqrt{62}}}\]
\[\overrightarrow{|a|} = \sqrt{(3)^2 + (-3)^2 + (2)^2 }=\sqrt{(9\ +\ 9\ +\ 4}=\sqrt{22}\]
\[\overrightarrow{|b|} = \sqrt{(4)^2 + (-2)^2 + (1)^2 }=\sqrt{(16\ +\ 4\ +\ 1}=\sqrt{21}\]
\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{62}}{\sqrt{22}\sqrt{21}}\]
\[\boxed{sin\ \theta =\frac{\sqrt{62}}{\sqrt{22}\sqrt{21}}}\]
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