Question

Determine Dot products and angle between vectors.

a) $u·v$

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

$u=i+3j,v=4i-j$

Short answer

a. $u·v=1$

b. $\theta =86°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=i+3j\\ \overrightarrow{v}=4i-j\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

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

$\overrightarrow{u} .\overrightarrow{v} =(1\times 4)+(3\times -1)$

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

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

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=i+3j\\ \overrightarrow{v}=4i-j\end{array}$

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

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

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

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

$|\overrightarrow{v} |=\sqrt{{\left(4\right)}^2+{(-1)}^2}$

$|\overrightarrow{v} |=\sqrt{16+1}$

$|\overrightarrow{v} |=\sqrt{17}$

Hence,

$\cos \theta =\frac{1}{\sqrt{10}\times \sqrt{17}}$

$\cos \theta =\frac{1}{\sqrt{170}}$

$\theta =86°$