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. $u·v=-1$

b. $\theta =97°$

Step-by-step solution

a.Step 1. Given information.

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

Step 2. Determining the dot product.

The dot product of vectors

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

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

$\overrightarrow{u} .\overrightarrow{v} =3-4$

$\overrightarrow{u} .\overrightarrow{v} =-1$

b.Step 1. Given information.

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

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =-1$

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

$|\overrightarrow{u} |=\sqrt{3^2+{(-2)}^2}$

$|\overrightarrow{u} |=\sqrt{9+4}$

$|\overrightarrow{u} |=\sqrt{13}$

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

$|\overrightarrow{v} |=\sqrt{1+4}$

$|\overrightarrow{v} |=\sqrt{5}$

Hence,

$\cos \theta =\frac{-1}{\sqrt{13}\times \sqrt{5}}$

$\cos \theta =\frac{-1}{\sqrt{65}}$

$\theta ={\cos }^{-1}\left(\frac{-1}{\sqrt{65}}\right)$

$\theta =97°$