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Question
Show that the vector \[\hat{i} + \hat{j} + \hat{k}\] is equally inclined to the coordinate axes.
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Solution
\[\text{ Let } \theta_1 \text{ be the angle between } \vec{a} \text{ and } x - axis.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{i} .........................\text{ (Because } \ \hat{i} \text{ is the unit vector along }x-axis)\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 1 + 0 + 0 = 1\]
\[\cos \theta_1 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta_1 = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 1 \right)\]
\[\]
\[\text{ Let } \theta_2 \text{ be the angle between } \vec{a} \ \text{ and } y - \text{ axis }\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{j} ........................... \text { (Because } \hat{j}\ \text{ is the unit vector along }y-\text{ axis) }\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 + 1 + 0 = 1\]
\[\cos \theta_2 = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta_2 = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 2 \right)\]
\[\text{ Let } \theta_3 \text{ be the angle between } \vec{a} \text{ and } z - \text{ axis }.\]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[ \vec{b} = \hat{k}.................................. { (Because } \ \hat{k}\ \text{ is the unit vector along }z-\text{ axis })\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2} = \sqrt{1} = 1\]
\[ \vec{a} . \vec{b} = 0 + 0 + 1 = 1\]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{1}{\left( \sqrt{3} \right)\left( 1 \right)} = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{3}} \right) . . . \left( 3 \right)\]
\[\text{ From } (1), (2) \text{ and } (3), \text{ the given vector is equally inclined to the coordinate axes }.\]
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