Determine whether u and v are orthogonal, parallel, or neither. u= (4/3,3/2). v=(-8,-9)
\[u=4/3+j3/2 , v=-8-j9 \]
Now this is a very interesting question: How to know whether two vectors are orhtogonal (which means perpendicular), parallel, or neither. remember our formula:\[\overrightarrow{u}*\overrightarrow{v}=|\overrightarrow{u}||\overrightarrow{v}|\cos\theta\]where theta is the angle between them. Now if the angle between them is theta=0 they point in the same direction, and hence are parallel. That makes the for cos(0)=1, so we get\[\overrightarrow{u}*\overrightarrow{v}=|\overrightarrow{u}||\overrightarrow{v}|\cos\theta=|\overrightarrow{u}||\overrightarrow{v}|\cos(0)=|\overrightarrow{u}||\overrightarrow{v}|\]so when the dot product is equal to the product of the magnitudes of the two vectors they are parallel.
what about perpendicular? then theta=pi/2 and we get cos(pi/2)=0, hence our formula becomes\[\overrightarrow{u}*\overrightarrow{v}=|\overrightarrow{u}||\overrightarrow{v}|\cos(\frac{\pi}{2})=|\overrightarrow{u}||\overrightarrow{v}|(0)=0\]so if we do the dot product and get zero we find out that the vectors are orthogonal (perpendicular).
If we check both of these facts and neither are true, then they are neither perpendicular nor parallel.
when i did the dot product they were neither the same and did not equal zero.
let me check...
I seem to get neither as well
awesome! thanks so much :)
yw!
wait I may be wrong, I think they are parallel...
yeahh i got that one wrong. lol its okay
Oh, only one chance? I'll remember to double-check next time :(
its perfectly fine! yeah just once chance
dang sorry, what I said above is true though, that's how I found out. Arithmetic error...
its alright! haha still thanks for the help
welcome
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