Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <6, -1>, v = <7, -4>

20.3°

10.2°

0.2°

30.3°

Answer 1

first find the scalar product of the 2 vectors
can you do this?

Answer 2

i do not remember well

Answer 3

$$
cos(\theta)\cfrac{|u|\times |v|}{||u||\times ||v||}
$$

Answer 4

its also called the dot product

u.v = 6*7 + -1*-4
= 42 + 4 = 46

you find the angle (x) using the formula

cos x = u.v
----
|u|.|v|

where |u| and |v| are the magnitudes of the 2 vectors

Answer 5

$$
\color{blue}{|u|\times |v|} \implies < a, b > \times < c , d >\\
\implies (a\times c) + (b \times d)\\
\color{blue}{||u|| \times ||v||} \implies \text{magnitudes of each multiplied}\\
\implies ||u|| = \sqrt{a^2+b^2}\\
\implies ||v|| = \sqrt{c^2+d^2}
$$

Answer 6

ok thanks i got it

Answer 7

$$
cos(\theta)=\cfrac{|u|\times |v|}{||u||\times ||v||}
$$
so you get
cos() = SOMEVALUE
arcsin both sides, to get the angle :)