Question

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

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

Answer 1

take the dot product.

Answer 2

so
<20, 12>?

Answer 3

if a and b are two vectors ,then
\[vector a. vector b=\left| a \right|\left| b \right|\cos \theta \]
where theta is the angle between the vectors.

Answer 4

so how would I start that?

Answer 5

@surjithayer `\(a\cdot b = ....\)`

Answer 6

\(\overline{a}\cdot \overline{b}=||\overline{a}|| \ ||\overline{b}||\cos(\theta)\)

Answer 7

:)

Answer 8

u=-5 i-4 j
v=-4 i-3 j
\[u.v=\left| u \right|\left| v \right|\cos \theta \]
\[u.v=(-5i-4j).(-4i-3j)=(-5)(-4)+(-4)(-3)=20+12=32\]
\[\left| u \right|=\sqrt{\left( -5 \right)^2+\left( -4)^2 \right)}=\sqrt{25+16}=\sqrt{41}\]
similarly find \[\left| v \right|\] and solve by the above formula.

Answer 9

What answer did you get? because i got sqrt of 37

Answer 10

I know it was wrong.