anonymous
  • anonymous
Find the unit vector in the direction of u, and write your answer in component form. Let u = <-8, -9>. Find 7u.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Please help !!
anonymous
  • anonymous
@amistre64 please help
anonymous
  • anonymous
@phi @ganeshie8 please help

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zzr0ck3r
  • zzr0ck3r
do you know how to find the magnitude of u?
zzr0ck3r
  • zzr0ck3r
@Andreshoyos23 ?
anonymous
  • anonymous
no
anonymous
  • anonymous
but i belive i figured it out @zzr0ck3r
zzr0ck3r
  • zzr0ck3r
if you have a vector (a,b) then the magnitude is \[\sqrt{a^2+b^2}\]
anonymous
  • anonymous
is it <-56, -63>
zzr0ck3r
  • zzr0ck3r
that is the answer to the second part
zzr0ck3r
  • zzr0ck3r
but the first part wants you to find the unit vector
anonymous
  • anonymous
the magnitude is squared and rooted?
anonymous
  • anonymous
meaning if i have the two numbers -8 and -9 you must square and muliply by two
zzr0ck3r
  • zzr0ck3r
\[given \space \\\text{then the unit vector is}\\\frac{1}{\sqrt{a^2+b^2}}*=<\frac{a}{\sqrt{a^2+b^2}},\frac{b}{\sqrt{a^2+b^2}}>\]
anonymous
  • anonymous
ah !!!! ok :D i see

Looking for something else?

Not the answer you are looking for? Search for more explanations.