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Section 5-3 : Dot Product

5. Determine the angle between \(\vec a = \vec i + 3\vec j - 2\vec k\) and \(\vec b = \left\langle { - 9,1, - 5} \right\rangle \).

Show Solution

Not really a whole lot to do here. All we really need to do is rewrite the formula from the geometric interpretation of the dot product as,

\[\cos \theta = \frac{{\vec a\centerdot \vec b}}{{\left\| {\vec a} \right\|\,\,\left\| {\vec b} \right\|}}\]

This will allow us to quickly determine the angle between the two vectors.

We’ll first need the following quantities (we’ll leave it to you to verify the arithmetic involved in these computations….).

\[\vec a\centerdot \vec b = 4\hspace{0.25in}\hspace{0.25in}\left\| {\vec a} \right\| = \sqrt {14} \hspace{0.25in}\hspace{0.25in}\left\| {\vec b} \right\| = \sqrt {107} \]

The angle between the vectors is then,

\[\cos \theta = \frac{4}{{\sqrt {14} \,\sqrt {107} }} = 0.1033\hspace{0.25in} \Rightarrow \hspace{0.25in}\theta = {\cos ^{ - 1}}\left( {0.1034} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{1.4673\,\,{\rm{radians}}}}\]