Thread: vectors and particles

Two particles R and S move so that their position vector, r and s respectively, are given by:
r = (2-2cos(t)) i + (1+sin(t)) j and s = sin(t) i + 2cos(t) j

where t seconds (t > 0) is the time elapsed since the start of their motions.
a) Evaluate r (dot) s and hence deduce the exact time when the position vectors of the two particles are first at right angles.

Now, what i got so far is sin(t) + cos(t) = 0, but im not too sure where to go from there, or if i am even in the right ballpark.

Any help appreciated.

Originally Posted by scorpion007

Two particles R and S move so that their position vector, r and s respectively, are given by:
r = (2-2cos(t)) i + (1+sin(t)) j and s = sin(t) i + 2cos(t) j

where t seconds (t > 0) is the time elapsed since the start of their motions.
a) Evaluate r (dot) s and hence deduce the exact time when the position vectors of the two particles are first at right angles.

Now, what i got so far is sin(t) + cos(t) = 0, but im not too sure where to go from there, or if i am even in the right ballpark.

Any help appreciated.

The dot product simplifies down to:

r.s = 2 cos(t) + 2 sin(t).

Thus r and s are at right angles when r.s=0, so we are looking for the
smallest positive root of:

cos(t) + sin(t) = 0,

which is where you had got to.

Now to get an idea of where this root is your best approach is to sketch
some curves. The point we seek is where:

cos(t) = -sin(t)

if we sketch these it appears that the first positive root is near t=3pi/4,
substituting this into the equation confirms that this is in fact a solution.

RonL

thank you very much.. i kinda thought i had to graph it, i just thought there may be a simple algebraic way to solve it.

Originally Posted by scorpion007

thank you very much.. i kinda thought i had to graph it, i just thought there may be a simple algebraic way to solve it.

If you know a lot about the sin and cos functions behaviour you
can just right down the solution, but the way one gets that familiarity
is by sketching the curves (it also stops one making some silly mistakes
so its still worth doing).

RonL