Question

Determine Dot products and angle between vectors.

a) $u·v$

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

$u=-5j,v=-i-\sqrt{3}j$

Short answer

a. $u·v=5\sqrt{3}$

b. $\theta =30°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=-5j\\ \overrightarrow{v}=-i-\sqrt{3}j\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

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

$\overrightarrow{u} .\overrightarrow{v} =(0\times -1)+(-5\times -\sqrt{3})$

$\overrightarrow{u} .\overrightarrow{v} =5\sqrt{3}$

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=-5j\\ \overrightarrow{v}=-i-\sqrt{3}j\end{array}$

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =5\sqrt{3}$

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

$|\overrightarrow{u} |=\sqrt{5^2}$

$|\overrightarrow{u} |=5$

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

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

$|\overrightarrow{v} |=2$

Hence,

$\cos \theta =\frac{5\sqrt{3}}{5\times 2}$

$\cos \theta =\frac{\sqrt{3}}{2}$

$\theta =30°$