Vector Help

[Borisan]

If anyone here is good with Calculus, please help me out. I have been trying to figure this problem out for a while now, but I am just stumped.

The problem states:

Find the vector v with the given magnitude and the same direction as u.

||v|| = 2 <---- magnitude

u = < sqrt3, 3 >


Please help! Or atleast give me a hint..

[Semitry]

Do you know how to find the direction of u? I.E. the angle from the origin ?

[Borisan]

No, I don't. : (

[Semitry]

Do you remember trig and right triangles?

if not brush up

http://www.themathpage.com/aTrig/tri...-triangles.htm

or you're fucked.

Once you have that down, just draw u with base sqrt(3) and height 3, find the direction from the origin. You know v's magnitude, ie hypotenuse and it's direction you can find it's base and height from that.

[Alexel]

If anyone here is good with Calculus, please help me out. I have been trying to figure this problem out for a while now, but I am just stumped.

The problem states:

Find the vector v with the given magnitude and the same direction as u.

||v|| = 2 <---- magnitude

u = < sqrt3, 3 >


Please help! Or atleast give me a hint..

sqrt((x*sqrt(3))^2 + (x*3)^2) = 2 -> solve for x

your new vector = <x*sqrt3, x*3>

I think this is right, but it's been a while, also this seems to be a pretty basic problem, if you can't figure this one out you should really spend some more time on this subject, or find an easier class.

[Jefe]

Please help! Or atleast give me a hint..

http://img120.imageshack.us/img120/9266/vectortd3.jpg

For two dimensional vector a = (a1, a2), therefore, |a| = sqrt(a1^2 + a2^2). I guess I should mention a1 = distance along x-axis, and a2 y-axis (just replace a for u in the pic above, too lazy to open PS).

[Charla]

first, find ||u||

That's sqrt((sqrt(3)^2) + 3^2) = sqrt(3+9) = sqrt(12)

Then, divide u itself by its own magnitude, to give you its unit vector (hint: if you don't know what a unit vector is, open the book again cause it's important)

That's <sqrt(3)/sqrt(12), 3/sqrt(12)> = <sqrt(3/12), sqrt(9)/sqrt(12)> = <sqrt(1/4), sqrt(3/4)> = <1/2, sqrt(3)/2> (in case you were wondering, this is easier to write than to type).

Now, just multiply that by the magnitude you wanted in the first place, which is 2. You get <1, sqrt(3)>.

Now, to double-check this, just find the magnitude of your result and make sure it's 2. So sqrt(1^2 + sqrt(3)^2) = sqrt(1+3) = sqrt(4) = duh.

Protip: this took me about 20 seconds to do in my head. It should be that easy for you by the time you graduate.

[aers]

Charla is correct. Are you taking Calc C? This is basic stuff that's the fundamentals for the entire class so you should really keep it memorized >_>

[Borisan]

Thank you all so much! Thats the only problem out of the 30 problems I couldn't do in the section..Finally got it!