Question

what is the angle that forms from the two vectors v= <-12,5> w=<3,10>

Answer 1

You have the following vectors:

$\begin{gathered} v=-12x+5y \\ w=3x+10y \end{gathered}$

By definition, the Dot product of two vectors

$\begin{gathered} a=hx+ry \\ b=mx+py \\ \end{gathered}$

is the following:

$a\cdot b=h\cdot m+r\cdot p$

Then you can calculate the Dot product with the vectors given in the exercise:

$\begin{gathered} v\cdot w=(-12)(3)+(5)(10) \\ v\cdot w=-36+50 \\ v\cdot w=14 \end{gathered}$

The Dot product between two vectors can be also written as:

$v\cdot w=|v|\cdot|w|\cdot\cos \alpha$

Where α is the angle between the vectors.

Now, in order to calculate the angle, you need to solve for the angle α:

$\begin{gathered} \cos \alpha=\frac{v\cdot w}{|v|\cdot|w|} \\ \\ \alpha=\cos ^{-1}(\frac{v\cdot w}{|v|\cdot|w|}) \end{gathered}$

You know that

$v\cdot w=14$

So now you need to find |v| amd |w|.

By definition:

$\begin{gathered} |v|=\sqrt[]{(-12)^2+(5)^2}=13 \\ \\ |w|=\sqrt[]{(3)^2+(10)^2}=10.44 \end{gathered}$

Knowing these values, you can calculate the angle:

$\begin{gathered} \alpha=\cos ^{-1}(\frac{v\cdot w}{|v|\cdot|w|}) \\ \\ \alpha=\cos ^{-1}(\frac{14}{13\cdot10.44}) \\ \\ \alpha\approx84.1\degree \end{gathered}$

The answer is:

$84.1\degree$