finding horizintal and vertical components of a vector

[icelated]

i have a problem and i cant figure it out.

I am trying to find the horizintal and vertical components of the vector u that starts at the point on the coordinate plane p = (-1,4)
and ends at the point Q = (3,2)

would i draw the vector on a plane then make a triangle?
i am thinking i would use these

x= rcos theta
y = rsin theta

so i would need to find r?
r = square root x^2 + y^2

and theta?
theta = arctan y/x

problem is i have two x,y
am i close?

 

[tkhunny]

It is your responsibility to put the angle in the right quadrant. If you get -14 degrees, it is likely something is wrong with a triangle solution. These things are periodic. Pick a different answer - one that makes sense.

 

[soroban]

Hello, icelated!

i have a problem and i cant figure it out. . Really?

I am trying to find the horizintal and vertical components of the vector \(\displaystyle \vec u\)
. . \(\displaystyle \text{that starts at }P(-1,4)\text{ and ends at }Q(3,2).\)

\(\displaystyle \text{Going from }P\text{ to }Q\text{, we move }3-(-1) \:=\:4 \text{ units horizontally}\)
. . \(\displaystyle \text{and }2-4 \:=\:-2\text{ units vertically.}\)

\(\displaystyle \text{The components of }\vec u\text{ are: }\:\langle 4,\,-2\rangle\)