Question

Determine Dot products and angle between vectors.

a) $u·v$

b) The angle between u and v to the nearest degree.

Short answer

a. $u·v$

b. $\theta =45°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=〈2,7〉\\ \overrightarrow{v}=〈3,1〉\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

$\overrightarrow{u} .\overrightarrow{v} =a_1{b}_1+a_2{b}_2$

$\overrightarrow{u} .\overrightarrow{v} =(2\times 3)+(7\times 1)$

$\overrightarrow{u} .\overrightarrow{v} =6+7$

$\overrightarrow{u} .\overrightarrow{v} =13$

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=〈2,7〉\\ \overrightarrow{v}=〈3,1〉\end{array}$

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =13$

Now, $\cos \theta =\frac{\overrightarrow{u} ·\overrightarrow{v} }{|\overrightarrow{u} ||\overrightarrow{v} |}$

$|\overrightarrow{u} |=\sqrt{2^2+7^2}$

$|\overrightarrow{u} |=\sqrt{4+49}$

$|\overrightarrow{u} |=\sqrt{53}$

$|\overrightarrow{v} |=\sqrt{3^2+1^2}$

$|\overrightarrow{v} |=\sqrt{9+1}$

$|\overrightarrow{v} |=\sqrt{10}$

Hence,

$\cos \theta =\frac{13}{\sqrt{53}\sqrt{10}}$

$\cos \theta =\frac{13}{\sqrt{530}}$

$\theta ={\cos }^{-1}\left(\frac{13}{\sqrt{530}}\right)$

$\theta =56°$