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

Part – A

\[1.\ Find\ the\ scalar\ product\ of\ \overrightarrow{i}+\overrightarrow{j}\ ,\ \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k}\ \hspace{10cm}\]

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

\[ \overrightarrow{a}.\overrightarrow{b}= 2\]
\[2.\ Show\ that\ the\ vectors\ \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}\ and\ – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k}\ are\ perpendicular\ to\ each\ other\ \hspace{10cm}\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k} \]
\[\overrightarrow{b}= – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}) .(- 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k})\]

= 2(3) + 3(2) – 2 (6)

= 6 + 6 -12

= 0

\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[2.\ Find\ the\ value\ of\ p\ if\ the\ vectors\ 2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k}\ and\ p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k} are\ perpendicular.\ \hspace{10cm}\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k} \]
\[ \overrightarrow{b}= p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k} \]
\[Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other \]
\[i.e\ \overrightarrow{a}.\overrightarrow{b}= 0\]
\[(2\overrightarrow{i}+ \overrightarrow{j}- 5\overrightarrow{k} ).(p\overrightarrow{i}+ 3\overrightarrow{j} – 2\overrightarrow{k}) = 0 .\]

2(p) + 1(3) – 5 (-2) = 0

2p + 3 +10 = 0

2p + 13 = 0

\[p = \frac{-13}{2}\]
\[3.\ If\ |\overrightarrow{a}| = 3,\ |\overrightarrow{b}|= 5\ and\ |\overrightarrow{a} × \overrightarrow{b}|=10,\ find\ the\ angle\ between\ \overrightarrow{a}\ and\ \overrightarrow{b}.\ \hspace{10cm}\]

Soln:

\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[ =\frac{10}{(3)(5)} = \frac{2}{3}\]
\[\theta\ = sin^{-1}(\frac{2}{3})\]
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Part –B

\[1.\ Find\ the\ projection\ of\ the\ vector\ 2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} on\ the\ vector\ \overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k}\ \hspace{10cm}\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}=\overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(2\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}).(\overrightarrow{i} – 2\overrightarrow{j} – 2\overrightarrow{k})}{\sqrt{(1)^2 + (-2)^2 + (-2)^2 }}\]
\[= \frac{2(1)+ 1(-2)- 2(-2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{2 – 2 + 4}{\sqrt{9}}\]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{4}{3}\]
\[2.\ Find\ the\ area\ of\ the\ parellelogram\ whose\ adjacent\ sides\ are\ \overrightarrow{i} + \overrightarrow{j} + 3\overrightarrow{k}\ and\ 2\overrightarrow{i} + \overrightarrow{j}+ 2\overrightarrow{k}.\ \hspace{10cm}\]

Soln:

\[\overrightarrow{a}=\overrightarrow{i} + \overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{b}= 2\overrightarrow{i} + \overrightarrow{j}+ 2\overrightarrow{k}\]
\[ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 3\\ 2 & 1 & 2\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 2 -3) -\overrightarrow{j}(3+1)+\overrightarrow{k}(3-0)\]
\[ = \overrightarrow{i}(-1) -\overrightarrow{j}(- 4)+\overrightarrow{k}(- 1)\]
\[ \overrightarrow{a}× \overrightarrow{b}= – \overrightarrow{i} + 4\overrightarrow{j} -\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-1)^2 + (4)^2 + (-1)^2 }=\sqrt{(1 + 16 +1 }=\sqrt{18}\]
\[ Area\ of \ parellelogram = 3\sqrt{2} sq.units\]
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Part –C

\[1.\ Prove\ that\ the\ vectors\ \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}\ and\ 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} are\ mutually\ perpendicular.\ \hspace{10cm}\]

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

\[\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{c}.\overrightarrow{a}= (7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}) .(\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k})\]

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

= 7 – 8 + 1

= 0

\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[2.\ Find\ the\ projection\ of\ the\ vector\ 8\overrightarrow{i}+ 4\overrightarrow{j}- 3\overrightarrow{k} on\ 2\overrightarrow{i} – 3 \overrightarrow{j} +2\overrightarrow{k}. Also\ find\ the\ angle\ between\ them \]

Soln:

\[\overrightarrow{a}= 8\overrightarrow{i}+ 4\overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i} – 3 \overrightarrow{j} +2\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (8\overrightarrow{i}+ 4\overrightarrow{j}- 3\overrightarrow{k}) .(2\overrightarrow{i} – 3 \overrightarrow{j} +2\overrightarrow{k})\]

=8 ( 2 ) + 4 ( -3) – 3 ( 2 )

=    16 – 12 – 6

=   – 2.

\[ \overrightarrow{a}.\overrightarrow{b}= = -2\]
\[\overrightarrow{|a|} = \sqrt{(8)^2 + (4)^2 + (-3)^2 }=\sqrt{(64 + 16 +9 }=\sqrt{89}\]
\[\overrightarrow{|b|} = \sqrt{(2)^2 + (-3)^2 + (2)^2 }=\sqrt{(4 + 9 + 4 }=\sqrt{17}\]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} = \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}} = \frac{-2}{ \sqrt{17}}\]
\[\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[= \frac{-2}{\sqrt{89}\sqrt{17}}\]
\[\theta = \cos ^{-1} ( \frac{-2}{\sqrt{89}\sqrt{17}})\]
\[3.\ Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ 5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k}\, \overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k} and\ -\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}.\]

Soln:

\[\overrightarrow{OA}= 5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k}\]
\[\overrightarrow{OB}= \overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}\]
\[\overrightarrow{OC}=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}- (5\overrightarrow{i}+2\overrightarrow{j} + 4\overrightarrow{k})\]
\[=\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k}- 5\overrightarrow{i}-2\overrightarrow{j} – 4\overrightarrow{k})\]
\[\overrightarrow{AB}= -4\overrightarrow{i} + \overrightarrow{j} – 2\overrightarrow{k}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}- (\overrightarrow{i} +3\overrightarrow{j}+ 2\overrightarrow{k})\]
\[=-\overrightarrow{i} – \overrightarrow{j}+\overrightarrow{k}- \overrightarrow{i} -3\overrightarrow{j} – 2\overrightarrow{k})\]
\[\overrightarrow{BC}= -2\overrightarrow{i} – 4\overrightarrow{j} – \overrightarrow{k}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -4 & 1 & -2\\ -2 & -4 & -1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -1 – 8) -\overrightarrow{j}(4 – 4)+\overrightarrow{k}(16 + 2)\]
\[ = \overrightarrow{i}(-9) – 0\overrightarrow{j}(-8)+\overrightarrow{k}(18)\]
\[\overrightarrow{AB}× \overrightarrow{BC}= -9\overrightarrow{i}+18\overrightarrow{k}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(9 )^2 + (18)^2}=\sqrt{(81 + 324 }=\sqrt{405}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[=\frac{\sqrt{405}}{2}\ sq. units\]
\[4.\ Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ \overrightarrow{i} – \overrightarrow{j}+ 3\overrightarrow{k} and\ 2\overrightarrow{i}+ 3\overrightarrow{j} -\overrightarrow{k}.\ \hspace{10cm}\]\[Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\ \hspace{10cm}\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i} – \overrightarrow{j}+ 3\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}+ 3\overrightarrow{j} -\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & -1 & 3\\ 2 & 3 & -1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 1 – 9) -\overrightarrow{j}(-1 – 6)+\overrightarrow{k}(3 + 2)\]
\[ = \overrightarrow{i}(-8) -\overrightarrow{j}(-7)+\overrightarrow{k}(5)\]
\[\overrightarrow{a}× \overrightarrow{b}= -8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-8 )^2 + (-7)^2 + (5)^2 }=\sqrt{(64 + 49 + 25}=\sqrt{138}\]
\[ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{-8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}}{\sqrt{138}} \]
\[ n^\wedge = \frac{-8\overrightarrow{i} + 7\overrightarrow{j}+5\overrightarrow{k}}{\sqrt{138}} \]
\[\overrightarrow{|a|} = \sqrt{(1)^2 + (-1)^2 + (3)^2 }=\sqrt{(1 + 1 + 9 }=\sqrt{11}\]
\[\overrightarrow{|b|} = \sqrt{(2)^2 + (3)^2 + (-1)^2 }=\sqrt{(4 + 9 + 1 }=\sqrt{14}\]
\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{138}}{\sqrt{11}\sqrt{14}}\]
\[\ sin\ \theta =\frac{\sqrt{138}}{\sqrt{11}\sqrt{14}}\]
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