Question

Determine Dot products and angle between vectors.

a) $u·v$

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

$u=3i+4j,v=-2i-j$

Short answer

a. $u·v=-10$

b. $\theta =86°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=3i+4j\\ \overrightarrow{v}=-2i-j\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

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

$\overrightarrow{u} .\overrightarrow{v} =(3\times -2)+(4\times -1)$

$\overrightarrow{u} .\overrightarrow{v} =-6-4$

$\overrightarrow{u} .\overrightarrow{v} =-10$

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=3i+4j\\ \overrightarrow{v}=-2i-j\end{array}$

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =-10$

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

$|\overrightarrow{u} |=\sqrt{3^2+4^2}$

$|\overrightarrow{u} |=\sqrt{25}$

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

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

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

Hence,

$\cos \theta =\frac{-10}{5\times \sqrt{5}}$

$\cos \theta =\frac{-2}{\sqrt{5}}$