Question

Find the angle between the given vectors to the nearest tenth of a degree. u = <6, 4>, v = <7, 5>

Answer 1

Answer: 1.8°

Step-by-step explanation:

To calculate the angle between the vectors u and v we use the formula of the dot product.

The dot product between two vecotores is:

$u\ *\ v = |u||v|*cosx$

Where x is the angle between the vectors

As we know the components of both vectors, we calculate the dot product by multiplying the components of both vectors

$u=6i + 4j\\v=7i +5j$

Then:

$u\ *\ v = 6*7 + 4*5$

$u\ *\ v = 42 + 20$

$u\ *\ v =62$

Now we calculate the magnitudes of both vectors

$|u|=\sqrt{6^2 + 4^2}\\\\|u|=2\sqrt{13}$

$|v|=\sqrt{7^2 +5^2}\\\\|v|=\sqrt{74}$

Then:

$62 = 2\sqrt{13}*\sqrt{74}*cosx$

Now we solve the equation for x

$62 = [tex]cosx=\frac{62}{2\sqrt{13}*\sqrt{74}}\\\\x=arcos(\frac{62}{2\sqrt{13}*\sqrt{74}})\\\\x=1.8\°$