Question

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

Question. Is there no picture here?

Answer 2

no picture

Answer 3

vector u=-5i+-4j
vector v=-4i-3j
\[If \theta is the \angle \between the vectors,then\]
\[\cos \theta=\frac{ u.v }{ \left| u \right|\left| v \right| }=\frac{ \left( -5i-4j \right).\left( -4i-3j \right) }{\sqrt{\left( 5 \right)^{2}+\left( -4 \right)^{2}} \sqrt{\left( -4\right)^{2} +\left( -3 \right)^{2}} }\]
\[\cos \theta=\frac{ \left( -5 \right)\left( -4 \right)+\left( -4 \right)\left( -3 \right) }{ 5\sqrt{41} }\]
\[\cos \theta=\frac{ 20+12 }{5\sqrt{41} }=\frac{ 32\sqrt{41} }{ 205 }\]
now you can calculate angle by calculator or log tables.

Answer 4

Well he has a picture. A very big picture o:

Answer 5

Which I cannot help with I haven't taken geometry in 2 years I am sorry ;-;

Answer 6

so the answer would be 0.9? @surjithayer

Answer 7

\[if \cos \theta=0.9,then calculate \cos^{-1} \left( 0.9 \right)\]

Answer 8

ok thank you

Answer 9

yw