Question

Determine Dot products and angle between vectors.

a) $u·v$

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

$u=2i+j,v=3i-2j$

Short answer

a. $u·v=4$

b. $\theta =60°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=2i+j\\ \overrightarrow{v}=3i-2j\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

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

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

$\overrightarrow{u} .\overrightarrow{v} =6-2$

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

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=2i+j\\ \overrightarrow{v}=3i-2j\end{array}$

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =4$

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

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

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

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

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

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

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

Hence,

$\cos \theta =\frac{4}{\sqrt{13}\times \sqrt{5}}$

$\cos \theta =\frac{4}{\sqrt{65}}$

$\theta ={\cos }^{-1}\left(\frac{4}{\sqrt{65}}\right)$

$\theta =60°$