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. $u·v=-12$

b. $\theta =180°$

Step-by-step solution

a.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=〈-6,6〉\\ \overrightarrow{v}=〈1,-1〉\end{array}$

Step 2. Determining the dot product.

The dot product of vectors

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

$\overrightarrow{u} .\overrightarrow{v} =(-6\times 1)+(6\times -1)$

$\overrightarrow{u} .\overrightarrow{v} =-6+(-6)$

$\overrightarrow{u} .\overrightarrow{v} =-12$

b.Step 1. Given information.

$\begin{array}{l}\overrightarrow{u}=〈-6,6〉\\ \overrightarrow{v}=〈1,-1〉\end{array}$

also from part (a) $\overrightarrow{u} ·\overrightarrow{v} =-12$

Now,

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

$|\overrightarrow{u} |=\sqrt{-6^2+6^2}$

$|\overrightarrow{u} |=\sqrt{36+36}$

$|\overrightarrow{u} |=6\sqrt{2}$

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

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

$|\overrightarrow{v} |=\sqrt{2}$

Hence,

$\cos \theta =\frac{-12}{6\sqrt{2}\times \sqrt{2}}$

$\cos \theta =\frac{-12}{12}$

$\theta ={\cos }^{-1}-1$

$\theta =180°$