# N – 2.1 – Vector Introduction – Exercise Problems with Solutions

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \overrightarrow{a}= 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k},\ \color {red} {find\ 3\overrightarrow{a}\ +\ \overrightarrow{b}}\ \hspace{10cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{a}= 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}$
$3\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 3(3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k})\ +\ \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}$
$=\ 9\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 3\overrightarrow{k}\ +\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}$
$3\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 10\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{5cm}$

$\color {purple} {2\ .}\ \overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k},\ \color {red} {find\ 2\overrightarrow{a}\ +\ 3\overrightarrow{b}}\ \hspace{10cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k}$
$2\overrightarrow{a}\ +\ 3\overrightarrow{b}= 2(2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}) + 3 ( 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k})$
$=4\overrightarrow{i}\ + 6\overrightarrow{j} + 2\overrightarrow{k} + 9\overrightarrow{i}\ – 3\overrightarrow{j} + 3\overrightarrow{k}$
$2\overrightarrow{a}\ +\ 3\overrightarrow{b}= 13\overrightarrow{i}\ + 3\overrightarrow{j} + 5\overrightarrow{k}\ \hspace{5cm}$

$\color {purple} {3\ .}\ If\ position\ vectors\ of\ the\ points\ A\ and\ B\ are\ 2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k}\ and\ 5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k},\ \color {red} {find\ \overrightarrow{|AB|}}\ \hspace{20cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{OA}= 2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{OB}= 5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k} – (2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k})$
$=5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k} – 2\overrightarrow{i}\ +\overrightarrow{j} – 3\overrightarrow{k}$
$\overrightarrow{AB}= 3\overrightarrow{i} + 2\overrightarrow{j} – 5\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(3)^2 + (2)^2 +(-5)^2 }$
$= \sqrt{(9 + 4 +25 }$
$= \sqrt{38}$

$\color {purple} {4\ .}\ \color {red} {Find\ the\ unit\ vector\ along\ the\ vector}\ 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}\ \hspace{20cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{a}= 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2+(-1)^2 }$
$= \sqrt{(4 + 1 +1}$
$=\sqrt{6}$
$\overrightarrow{|a|}=\sqrt{6}$
$Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}}{\sqrt{6}}$

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {5\ .}\ If\ the\ position\ vector\ of\ the\ points\ A\ and\ B\ are\ \hspace{15cm}$$\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k},\ \hspace{12cm}$$\color {red} {find\ \overrightarrow{|AB|}\ ,\ Also\ find\ the\ direction\ ratio\ of\ \overrightarrow{AB}}\ \hspace{10cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$\overrightarrow{OA}\ =\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{OB}\ =\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}- (\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})$
$=\ 3\overrightarrow{i}\ +\ 2 \overrightarrow{j}\ +\ 3\overrightarrow{k}\ -\ \overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{|AB|} = \sqrt{(2)^2 + (3)^2 +(2)^2 }$
$= \sqrt{(4\ +\ 9\ +\ 4})$
$\boxed{\overrightarrow{|AB|}\ = \sqrt{17}}$
$Direction\ ratios\ are\ 2, 3, 2$

$\color {purple} {6\ .}\ \color {red} {Find\ the\ Modulus\ and\ Direction\ cosines}\ \hspace{15cm}$$of\ the\ vector\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{12cm}$
$\color {blue} {Soln:}\ Given\ \hspace{20cm}$
$Let\ \overrightarrow{a}\ =\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$r =\overrightarrow{|a|} = \sqrt{(2)^2 + (3)^2 + (4)^2 }$
$= \sqrt{(4\ +\ 9\ +\ 16}$
$r =\sqrt{29}$
$Direction\ cosines\ are \frac{2}{\sqrt{29}}\ , \frac{3}{\sqrt{29}}\ , \frac{4}{\sqrt{29}}$

$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {7\ .}\ \color {red} {\ Show\ that}\ the\ points\ whose\ position\ vectors\ \hspace{15cm}$$\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ \overrightarrow{k},\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}\ and\ -\ 7 \overrightarrow{j}\ -\ 5\overrightarrow{k}\ \color {red} {are\ collinear}\ \hspace{5cm}$
$\color{blue} {Soln\ :}\ Given\ \hspace {18cm}$
$\overrightarrow{OA}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{OB}= 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{OC}\ =\ -\ 7 \overrightarrow{j}\ -\ 5\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – (\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ \overrightarrow{k})$
$=2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – \overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ \overrightarrow{k}$
$\overrightarrow{AB}\ = \overrightarrow{i}\ +\ 5\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ -\ 7 \overrightarrow{j}\ -\ 5\overrightarrow{k}\ – (2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 3\overrightarrow{k})$
$=\ -\ 7 \overrightarrow{j}\ -\ 5\overrightarrow{k}\ – 2\overrightarrow{i}\ – 3\overrightarrow{j}\ – 3\overrightarrow{k}$
$\overrightarrow{BC}\ =\ -\ 2\ \overrightarrow{i}\ – 10\ \overrightarrow{j}\ -\ 8\overrightarrow{k}$
$=\ -\ 2( \overrightarrow{i}\ +\ 5\overrightarrow{j}\ +\ 4\overrightarrow{k})$
$\overrightarrow{BC}\ =\ -\ 2\ \overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$

$\color {purple} {8\ .}\ \color {red} {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}\ \color {red} {form\ an\ equilateral\ triangle}$
$\color{blue} {Soln\ :}\ Given\ \hspace {18cm}$
$\overrightarrow{OA}\ =\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}$
$\overrightarrow{OB}\ =\ 3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{OC}\ =\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\ 3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 2\overrightarrow{k}\ – (2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k})$
$=\ 3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 2\overrightarrow{k}\ -\ 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ -\ 4\overrightarrow{k}$
$\overrightarrow{AB}\ =\ \overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(1)^2\ +\ (1)^2\ +\ (-2)^2 }$
$= \sqrt{(1\ +\ 1\ +\ 4})$
$AB = \sqrt{6}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ -\ (3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 2\overrightarrow{k})$
$=\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – 3\overrightarrow{i}\ – 4\overrightarrow{j}\ -\ 2\overrightarrow{k}$
$\overrightarrow{BC}= \overrightarrow{i} -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-2)^2 +(1)^2 }$
$= \sqrt{(1\ +\ 4\ +\ 1})$
$BC = \sqrt{6}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ -\ (2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k})$
$=\ 4\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ -\ 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ -\ 4\overrightarrow{k}$
$\overrightarrow{AC}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(2)^2 + (-1)^2 +(-1)^2 }$
$= \sqrt{(4\ +\ 1\ +\ 1}$
$AC = \sqrt{6}$
$AB = BC = AC = \sqrt{6}$
$\therefore\ The\ given\ triangle\ is\ an\ equilateral\ triangle$

$\color {purple} {9\ .}\ \color {red} { Prove\ that\ the\ points}\ \hspace{15cm}$$3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2 \overrightarrow{k}\ ,\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k} and\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ \color {red} {form\ an\ isosceles\ triangle}$
$Soln:\ Given\ \hspace{20cm}$
$\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}$
$\overrightarrow{OB}\ =\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}$
$\overrightarrow{OC}\ =\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})$
$=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(2)^2\ +\ (2)^2\ +\ (-1)^2 }$
$= \sqrt{(4\ +\ 4\ +\ 1)}$
$=\ \sqrt{9}$
$AB =\ 3$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k})$
$=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 5\overrightarrow{i}\ – \overrightarrow{j}\ +\ 3\overrightarrow{k}$
$\overrightarrow{BC}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 2\overrightarrow{k}$
$BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (- 2)^2 +(2)^2 }$
$= \sqrt{(1\ +\ 4\ +\ 4)}$
$=\ \sqrt{9}$
$BC =\ 3$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})$
$=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}$
$\overrightarrow{AC}\ =\ 3\overrightarrow{i}\ +\ \overrightarrow{k}$
$AC =\overrightarrow{|AC|} = \sqrt{(3)^2\ +\ (0)^2\ +\ (1)^2 }$
$= \sqrt{(9\ +\ 0\ +\ 1)}$
$AC = \sqrt{10}$
$\boxed{AB = BC\ \neq\ AC}$

The given triangle is an isosceles triangle.