Calculus II - Dot Product
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Section 11.3 : Dot Product

5. Determine the angle between a=i+3j2k and b=9,1,5.

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,

cosθ=abab

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….).

ab=4a=14b=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}}}}