N – 2.1 – Vector Introduction – Exercise Problems with Solutions

\[\underline{PART\ -\ A}\]
\[1.\ If\ \overrightarrow{a}= 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\, find\ 3\overrightarrow{a}\ +\ \overrightarrow{b}\ \hspace{10cm}\]
\[Soln:\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\]
\[3\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 3(3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k})\ +\ \overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\]
\[=\ 9\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 3\overrightarrow{k}\ +\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[3\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 10\overrightarrow{i}\ +\ 9\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{5cm}\]
\[2.\ If\ \overrightarrow{a}= 2\overrightarrow{i}\ + 3\overrightarrow{j} + \overrightarrow{k}\ and\ \overrightarrow{b}= 3\overrightarrow{i}\ – \overrightarrow{j} + \overrightarrow{k}\, find\ 2\overrightarrow{a}\ +\ 3\overrightarrow{b}\ \hspace{10cm}\]

Soln: Given

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

Sol.:     Given

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

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2+(-1)^2 }\]
\[= \sqrt{(4 + 1 +1}\]
\[=\sqrt{6}\]
\[\overrightarrow{|a|}=\sqrt{6}\]
\[Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{k}}{\sqrt{6}}\]
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\[\underline{PART\ -\ B}\]
\[5.\ If\ the\ position\ vector\ of\ the\ points\ A\ and\ B\ are\ \hspace{15cm}\]\[\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k},\ \hspace{12cm}\]\[find\ \overrightarrow{|AB|}\ ,\ Also\ find\ the\ direction\ ratio\ of\ \overrightarrow{AB}\ \hspace{10cm}\]
\[Soln:\ Given\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ 3\overrightarrow{k}- (\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})\]
\[=\ 3\overrightarrow{i}\ +\ 2 \overrightarrow{j}\ +\ 3\overrightarrow{k}\ -\ \overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{|AB|} = \sqrt{(2)^2 + (3)^2 +(2)^2 }\]
\[ = \sqrt{(4\ +\ 9\ +\ 4})\]
\[\boxed{\overrightarrow{|AB|}\ = \sqrt{17}}\]
\[ Direction\ ratios\ are\ 3, 4, -5 \]
\[6.\ Find\ the\ Modulus\ and\ Direction\ cosines\ \hspace{15cm}\]\[of\ the\ vector\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{12cm}\]
\[Soln:\ Given\ \hspace{20cm}\]
\[Let\ \overrightarrow{a}\ =\ 2\overrightarrow{i}\ + 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(2)^2 + (3)^2 + (4)^2 }\]
\[ = \sqrt{(4\ +\ 9\ +\ 16}\]
\[r =\sqrt{29}\]
\[Direction\ cosines\ are \frac{2}{\sqrt{29}}\ , \frac{3}{\sqrt{29}}\ , \frac{4}{\sqrt{29}} \]
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\[\underline{PART\ -\ C}\]
\[7.\ Prove\ that\ the\ points\ \hspace{15cm}\]
\[3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2 \overrightarrow{k}\ ,\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k} and\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ form\ an\ isosceles\ triangle\]
\[Soln:\ Given\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(2)^2\ +\ (2)^2\ +\ (-1)^2 }\]
\[ = \sqrt{(4\ +\ 4\ +\ 1)}\]
\[ =\ \sqrt{9}\]
\[AB =\ 3 \]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 5\overrightarrow{i}\ – \overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{BC}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (- 2)^2 +(2)^2 }\]
\[ = \sqrt{(1\ +\ 4\ +\ 4)}\]
\[ =\ \sqrt{9}\]
\[BC =\ 3\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ 3\overrightarrow{i}\ +\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(3)^2\ +\ (0)^2\ +\ (1)^2 }\]
\[ = \sqrt{(9\ +\ 0\ +\ 1)}\]
\[AC = \sqrt{10}\]
\[\boxed{AB = BC\ \neq\ AC}\]

The given triangle is an isosceles triangle.

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