Question

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

u = <2, -4>, v = <3, -8>

Answer 1

\(\bf \textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\
\theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\)

Answer 2

I used the formula cos(theta) = u * v / ||u|| ||v|| to get 38/ sqrt20 * sqrt73 and im not getting any of the listed answers, perhaps im plugging it into the calcu wrong?

Answer 3

hmm what did you get for the dot product?

Answer 4

38

Answer 5

one sec

Answer 6

Oh I didnt use cos^-1. Is the answer 6.0deg?

Answer 7

\(\bf \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\to \left(\cfrac{38}{\sqrt{(2)^2+(-4)^2}\cdot \sqrt{(3)^2+(-8)^2}}\right)
\\ \quad \\
\to \left(\cfrac{38}{\sqrt{20}\cdot \sqrt{73}}\right)\to \left(\cfrac{38}{\sqrt{1460}}\right)\quad thus\to \theta=cos^{-1}\left(\cfrac{38}{\sqrt{1460}}\right)\)

Answer 8

yeap 6.00900595 degrees, or \(6^o\)

Answer 9

Thanks, and Im also stuck on this:
Evaluate the expression
v*w
Given the vectors
r=<8,1,-6>; v=<6,7,-3); w=<-7,5,2>

Does r play any role into this
and would -27 be the wrong answer?