Physics, magnitude and direction, and xy components?

Question

Vector a is directed along the positive y-axis and has a magnitude of 3.0 square root, . vector b is directed along the negative x-axis and has magnitude 1.0 unit. what are the magnitude adn direction of vectors a and b. what are the x- and y- components of vectors b-a?

would you use the pathagorian theorum to solve the first question? how do you find the x and y components for vector b-a?

Answer 1

the first vector can be writen as 0\hat{x} + \sqrt{3}\hat{y}
the second -1\hat{x} + 0\hat{y}
where \hat{x} should be read as "x hat" ,which is the unit vector in the x direction.

now.... to get the magnitude of b -a ... an easier way to think about it is b + (-a)

so you draw out the triangle. you draw one of the vectors, then at the tip of it draw the other so you will create a right triangle with a base of \sqrt{3} and a hight of 1
.... or mathimaticly
your new vector is 1\hat{x} + \sqrt{3}\hat{y}


now you use pathagorist theory to find the length of the other side. (\sqrt{3})^2 + 1^2 = 4^2
so the magnitude of the third side is 2

to find the direction you have to use some trig. namely use the tangent of the oposit over the adjacent is equal to the angle between the new vector and the x axis
... theta = tan(\sqrt{3} /1)

and if you remeber your geomotry this is actualy a 30,60,90 triangel so the angle is 60 degrees.

I think i have solved this basicly to completion... feel free to email me if you have more questions.

Answer 2

properly, sounds such as you are trying to study a subject remember without examining it rather is stipulations speed is a vector volume , (cost and direction ) and that pointer is as long because it rather is cost,,,,,, pointing at some direction enable's say it rather is making an attitude A between it rather is direction and the x-axis so, it rather is "shadow" on an axis is it rather is element comparable to that axis for our case, | vector | / | / | / ) | /A ) |------------------- x-axis |------| it rather is x-element (the shadow) you may get the element trigonometrically cost * cos( attitude between the vector and the in touch axis ) in our case, if the vector's call is V ||V|| * cos(A) ||V|| skill the cost of V (norm V) sorry for the undesirable graph, yahoo solutions ought to upload some drawing facilitators

Answer 3

review here

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/vectors/u3l1b.html