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Question
Show that the vectors \[\vec{a} = \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right), \vec{b} = \frac{1}{7}\left( 3\hat{i} - 6 {j} + 2 \hat{k} \right), \vec{c} = \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 {k} \right)\] are mutually perpendicular unit vectors.
Solution
\[\left| \vec{a} \right| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1\]
\[\left| \vec{b} \right| = \frac{1}{7}\sqrt{3^2 + \left( - 6 \right)^2 + 2^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1\]
\[\left| \vec{c} \right| = \frac{1}{7}\sqrt{6^2 + 2^2 + \left( - 3 \right)^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1\]
\[\text{ And }\]
\[ \vec{a} . \vec{b} \]
\[ = \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 {k} \right) . \frac{1}{7}\left( 3 t{i} - 6 \hat{j} + 2 \hat{k} \right)\]
\[ = \frac{1}{49}\left( 6 - 18 + 12 \right)\]
\[ = 0\]
\[ \vec{b} . \vec{c} \]
\[ = \frac{1}{7}\left( 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \right) . \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \right)\]
\[ = \frac{1}{49}\left( 18 - 12 - 6 \right)\]
\[ = 0\]
\[ \vec{c} . \vec{a} \]
\[ = \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \right) . \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\]
\[ = \frac{1}{49}\left( 12 + 6 - 18 \right)\]
\[ = 0\]
\[So,\left| \vec{a} \right| = \left| \vec{b} \right| = \left| \vec{c} \right| = 1\text{ and } \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0\]
\[\text{ So }, \text{ the given vectors are mutually perpendicular unit vectors. }\]
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