1. ## vector dot product

Hello
I have a question which I am not happy with

Find the angle between a and b given

|a|= 4
|b|= 4
and
|a-b|= 7

I have drawn it out and just got myself confused.
Thank you

2. Magnitude of (a-b) is given as

$\displaystyle R^2 = a^2 + b^2 - 2abcos(\theta)$

Substitute the values and find θ.

3. Method I.
Let $\displaystyle \overrightarrow{a}=\overrightarrow{AB}, \ |\overrightarrow{a}|=|\overrightarrow{AB}|=AB=4$

$\displaystyle \overrightarrow{b}=\overrightarrow{AC}, \ |\overrightarrow{b}|=|\overrightarrow{AC}|=AC=4$

Then $\displaystyle \overrightarrow{a}-\overrightarrow{b}=\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CB}$

and $\displaystyle |\overrightarrow{a}-\overrightarrow{b}|=|\overrightarrow{CB}|=CB=7$

So, $\displaystyle \overrightarrow{a}, \ \overrightarrow{b}, \ \overrightarrow{a}-\overrightarrow{b}$ are the sides of the triangle $\displaystyle ABC$.

From the cosine law we have

$\displaystyle \cos A=\displaystyle\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=-\frac{17}{32}$

Method II
$\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2=(\overrightarrow{a}-\overrightarrow{b})^2=|\overrightarrow{a}|^2+|\ove rrightarrow{b}|^2-2\cdot\overrightarrow{a}\cdot\overrightarrow{b}$

Then $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=-\displaystyle\frac{17}{2}$

Now, $\displaystyle \cos\left(\widehat{\overrightarrow{a},\overrightar row{b}}\right)=\displaystyle\frac{\overrightarrow{ a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\cd ot|\overrightarrow{b}|}=-\frac{17}{32}$

4. Thanks to both
I shall have a go at this and come back if I am struggling