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Question
Find the unit vector in the direction of vector \[\overrightarrow{PQ} ,\]
where P and Q are the points (1, 2, 3) and (4, 5, 6).
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Solution
Let \[\vec{a}\] and \[\vec{b}\] are the position vectors of the points \[P\left( 1, 2, 3 \right)\] and \[Q\left( 4, 5, 6 \right)\] Then,
\[\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \vec{b} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}\]
So,
\[\overrightarrow{PQ} = \vec{b} - \vec{a} \]
\[ = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} \]
\[ = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}\]
Now,
\[\left| \overrightarrow{PQ} \right| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = 3\sqrt{3}\]
Therefore, Unit vector parallel to \[\vec{PQ}\] = \[\frac{\overrightarrow{PQ}}{\left| PQ \right|} = \frac{1}{3\sqrt{3}} \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right) = \frac{1}{\sqrt{3}}\left( \hat{i} + \hat{j} + \hat{k} \right)\]
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