Question

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

u = <-5, -4>, v = <-4, -3>

-9.1°
1.8°
0.9°
11.8°

Answer 1

you can draw them out.

Answer 2

or use a property of a vector like dot product.

Answer 3

yeah i can only get so far working it out then i get stuck

Answer 4

show us what you worked out so we can help you from there

Answer 5

\[\vec u \cdot \vec v = |\vec u||\vec v|\cos(\theta)\]\[\theta = \cos^{-1}\left(\frac{\vec u \cdot \vec v}{|\vec u||\vec v|}\right)\]

Answer 6

\[\vec u = \langle u_1~,~u_2\rangle~,~ \vec v=\langle v_1~,~v_2\rangle\]\[\vec u \cdot \vec v = u_1v_1 +u_2v_2\]

Answer 7

\[|\vec u| = \sqrt{(u_1)^2 +(u_2)^2}\]\[|\vec v| = \sqrt{(v_1)^2 + (v_2)^2}\]

Answer 8

Now you've got all you need, you can find the angle.

Answer 9

i get up to \[\theta=\cos^{-1} \left[\begin{matrix}(-5)(-4) + (-4)(-3) \\ \left| 6.40 \right| \left| 5 \right|\end{matrix}\right]\]

then my mind goes blank

Answer 10

@Jhannybean

Answer 11

\[\theta = \cos^{-1}\left(\frac{(-5)(-4)+(-4)(-3)}{\sqrt{41}\sqrt{25}}\right)\]\[\theta = \cos^{-1}\left(\frac{20+12}{5\sqrt{41}}\right)\]\[\theta = ~?\]

Answer 12

The answer is C... 0.9 degrees