Question

I can't get this.. Help?
Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

Answer 1

\[ if~ \theta~be~the~\angle~\between~the~vectors~u~and~v`then~u.v=\left| u \right|\left| v \right|\cos \theta \]

Answer 2

I have no idea what that means...

Answer 3

I can't get the ||u|| part.. What do those things mean?

Answer 4

u=8i+7j
v=9i+7j
\[u.v=\left( 8i+7j \right).\left( 9i+7j \right)=\left( 8 \right)\left( 9 \right)+\left( 7 \right)\left( 7\right)=72+49=121\]
\[\left| u \right|=\sqrt{8^2+7^2}=\sqrt{64+49}=\sqrt{113}=3\sqrt{13}\]
\[\left| v \right|=\sqrt{9^2+7^2}=\sqrt{81+49}=\sqrt{130}\]
\[121=3\sqrt{13}\sqrt{130}\cos \theta=3(13)\sqrt{10}\cos \theta ,\]
\[\cos \theta=\frac{ 121 }{39\sqrt{10} }=\frac{ 121\sqrt{10} }{ 390 }=?\]

Answer 5

.9811 Then do I do cos^-1(.9811)?

Answer 6

Nope.. That's not it.

Answer 7

@surjithayer

Answer 8

@surjithayer \[\sqrt{113}\ne3\sqrt{13}\]because \(9*13= 117, \text{ not } 113\)

Answer 9

Oh. -_-

Answer 10

YES YOU ARE CORRECT
\[\cos \theta=\frac{ 121 }{\sqrt{113}\sqrt{130} }=0.9983,\theta=\cos^{-1} 0.9983 =3.3 ~degree\]

Answer 11

That'll derail any of the computations downstream from there, of course...

Answer 12

sorry for that.

Answer 13

Okay, got it! You're fine, hun. We're only human. :3 Thank you! God bless you!

Answer 14

oh, good, you're available, I'm in the middle of cooking dinner, and was dreading fixing the problem :-)

Answer 15

by the way, if you want to make a degree symbol, use ^\circ

Answer 16

in the equation

Answer 17

or on the computer.

Answer 18

Are you talking to me or him?

Answer 19

I meant if you wanted to have something like \[3.3^\circ\]

\ [3.3^\circ\ ] (without the spaces) will do the trick

Answer 20

\[\left[ 3.5^\circ \right] \]

Answer 21

Oh, okay. :3 Thanks!