N – 2.1 – Vector Introduction – Exercise Problems with Solutions

Part – A

$1.\ If\ \overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k}\, find\ 2\overrightarrow{a}\ +\ 3\overrightarrow{b}\ \hspace{10cm}$

Soln: Given

$\overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k}$
$2\overrightarrow{a}\ +\ 3\overrightarrow{b}= 2(2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}) + 3 ( 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k})$
$=4\overrightarrow{i}\ + 6\overrightarrow{j} + 2\overrightarrow{k} + 9\overrightarrow{i}\ – 3\overrightarrow{j} + 3\overrightarrow{k}$
$2\overrightarrow{a}\ +\ 3\overrightarrow{b}= 13\overrightarrow{i}\ + 3\overrightarrow{j} + 5\overrightarrow{k}\ \hspace{5cm}$
$2.\ If\ position\ vectors\ of\ the\ points\ A\ and\ B\ are\ 2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k}\ and\ 5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k}\ find\ \overrightarrow{|AB|}\ \hspace{20cm}$

Sol.:     Given

$\overrightarrow{OA}= 2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{OB}= 5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k} – (2\overrightarrow{i}\ -\overrightarrow{j} + 3\overrightarrow{k})$
$=5\overrightarrow{i}\ + \overrightarrow{j} – 2\overrightarrow{k} – 2\overrightarrow{i}\ +\overrightarrow{j} – 3\overrightarrow{k}$
$\overrightarrow{AB}= 3\overrightarrow{i} + 2\overrightarrow{j} – 5\overrightarrow{k}$
$AB =\overrightarrow{|AB|} = \sqrt{(3)^2 + (2)^2 +(-5)^2 }$
$= \sqrt{(9 + 4 +25 }$
$= \sqrt{38}$
$3.\ Find\ the\ unit\ vector\ along\ the\ vector\ 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}\ \hspace{20cm}$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2+(-1)^2 }$
$= \sqrt{(4 + 1 +1}$
$=\sqrt{6}$
$\overrightarrow{|a|}=\sqrt{6}$
$Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}}{\sqrt{6}}$

Part –B

1. Show that the points whose position vectors

$2\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k},\ 3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}\ and\ -\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}\ are\ collinear\ \hspace{10cm}$

Soln:   Given

$\overrightarrow{OA}= 2\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k}$
$\overrightarrow{OB}= 3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{OC}= -\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}- (2\overrightarrow{i}\ – \overrightarrow{j} + 3\overrightarrow{k})$
$=3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k}- 2\overrightarrow{i}\ + \overrightarrow{j} – 3\overrightarrow{k}$
$\overrightarrow{AB}= \overrightarrow{i} – 4\overrightarrow{j} – 2\overrightarrow{k}$
$\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}$
$=-\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}- (3\overrightarrow{i}\ – 5\overrightarrow{j} + \overrightarrow{k})$
$=-\overrightarrow{i}\ +11 \overrightarrow{j}+ 9\overrightarrow{k}- 3\overrightarrow{i}\ + 5\overrightarrow{j} – \overrightarrow{k})$
$\overrightarrow{BC}= -4\overrightarrow{i} + 16\overrightarrow{j} + 8\overrightarrow{k}$
$\overrightarrow{BC}= -4(\overrightarrow{i} – 4\overrightarrow{j} – 2\overrightarrow{k})$
$\overrightarrow{BC} = -4\overrightarrow{AB}$
$\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear$

Part –C

1. Prove that the points
$2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k}, 3\overrightarrow{i}\ + 4\overrightarrow{j}+ 2\overrightarrow{k} and\ 4\overrightarrow{i}\ +2 \overrightarrow{j}+ 3\overrightarrow{k}\ form\ an\ equilateral\ triangle$

Soln: Given

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

The given triangle is an equilateral triangle.

2. Prove that the points whose position  vectors  are

$2\overrightarrow{i}\ – \overrightarrow{j}+ \overrightarrow{k}, \overrightarrow{i}\ – 3\overrightarrow{j} – 5\overrightarrow{k} and\ 3\overrightarrow{i}\ -4 \overrightarrow{j} – 4\overrightarrow{k}\ form\ a\ right\ angled\ triangle$

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.