Direction of Vectors

[Whatupdoc]

Vector A has components Ax=1.50cm, Ay =2.05cm
Vector B has components Bx=4.40cm, By=-3.65cm

Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)

Hi, can someone help me with this problem. i know that it's easy, but i dont remeber how to do it. thanks

[ExtravagantDreams]

Vectors add just like integers. Add the x components and the y components. Draw yourself a picture for it to make more sense.

Once you have your Cx and Cy, you can use your choise of arctan, arcsin with corresponding sides to find the angle.

[Whatupdoc]

Vectors add just like integers. Add the x components and the y components. Draw yourself a picture for it to make more sense.

Once you have your Cx and Cy, you can use your choise of arctan, arcsin with corresponding sides to find the angle.

yea i know how to add vectors, but i dont know what the question is asking me to do. "Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)" i'm very confused, what is the question asking me to do extactly? sorry, i really suck when it comes to physics.

[ehild]

yea i know how to add vectors, but i dont know what the question is asking me to do. "Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)" i'm very confused, what is the question asking me to do extactly?

You need to find the angle the vector C=A+B encloses with the x axis. Remember that
tan(\gamma ) = c_y/c_x
From this, you get gamma back, but decide if the angle is in the first, second, third or fourth quadrant.
If both components are positive
0 < \gamma < \pi/2
if the x component is negative and the y component is positive
\pi/2 < \gamma < \pi
If both components are negative
\pi <\gamma <3 \pi /2
if the x component is positive and the y component is negative
3\pi /2 < \gamma < 2 \pi

So calculate:
c_x = a_x + b_x and c_y = a_y + b_y
tan(\gamma ) = c_y/c_x
find gamma, you get an angle between 0 and \pi (or between 0 and 180^o ). Check if c_y is positive or negative. If it is negative add \pi (180^o ) to the value of the angle you got.


ehild

[ExtravagantDreams]

"Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)" i'm very confused, what is the question asking me to do extactly?

Not sure what class this is for but I remember for my first physics class we would write a vector as C [\theta deg. above positive x-axis]

[robphy]

yea i know how to add vectors, but i dont know what the question is asking me to do. "Find the Direction of Vector A + B. (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from the axis)" i'm very confused, what is the question asking me to do extactly? sorry, i really suck when it comes to physics.

This is actually a problem in mathematics. Don't blame Physics! :mad: :tongue2:

After doing the addition of vectors, as described in the other posts here, you are essentially converting from rectangular coordinates to polar coordinates.
This is probably in your trig or pre-calc text.

In physics, one merely uses that mathematical idea.

[Whatupdoc]

You need to find the angle the vector C=A+B encloses with the x axis. Remember that
tan(\gamma ) = c_y/c_x
From this, you get gamma back, but decide if the angle is in the first, second, third or fourth quadrant.
If both components are positive
0 < \gamma < \pi/2
if the x component is negative and the y component is positive
\pi/2 < \gamma < \pi
If both components are negative
\pi <\gamma <3 \pi /2
if the x component is positive and the y component is negative
3\pi /2 < \gamma < 2 \pi

So calculate:
c_x = a_x + b_x and c_y = a_y + b_y
tan(\gamma ) = c_y/c_x
find gamma, you get an angle between 0 and \pi (or between 0 and 180^o ). Check if c_y is positive or negative. If it is negative add \pi (180^o ) to the value of the angle you got.


ehild


ok here's what i got:
c_x = 5.9cm
c_y = -1.6cm

so...
(\gamma ) = -15.17291

the angle is negative, so i added 180 to it, which equals .... 164.827 <-- is that the answer?

[faust9]

Add 360, not 180. Remember, the A+B vector is in the 4th quad so to get the CCW angle from the x-axis you have go through the first three quads which is 270 degrees. And then you need to go an additional 75'ish degrees to find the angle from the x-axis to the vector. Or the easier route of 360+\gamma

[ehild]

ok here's what i got:
c_x = 5.9cm
c_y = -1.6cm

so...
(\gamma ) = -15.17291

the angle is negative, so i added 180 to it, which equals .... 164.827 <-- is that the answer?


Ooops...Sorry, I was wrong, faust9 is right.
The calculator returns an angle between -90^o and +90^o. Add 180 degrees if c_x <0 but 360 degrees if c_x>0 \mbox{ and } c_y<0.
So your angle is 344.8^o Remember, you have to check the sign of both components, it is not enough to check if the angle is negative or positive.

Try the vectors {1,1); (-1,1); (-1,-1); (1,-1). They make the angles with the positive x axis counter-clockvise 45^o, 135^o, 225^o, 315^0 . The tangents are: 1, -1, 1, -1. The function arctan returns 45, -45, 45 -45. You have to add 0, 180, 180, 360, respectively, to get back the original angles.

ehild