Lines and planes in space

[Guest hellokitty818]

So for my hw...i have the question

boats a position is given by x(t)=3-t, y(t)=2t-4 where the distance units are kilometers and the time units are hours. boats B's position is x(t)=4-3t, y(t)=3-2t

then i found part a and b...then for part c it asks "what is the angle between the paths of the boats?''

but now i am stuck...i plugged it into the cosine equation cosx=(AdotB)/(A)(B)

(() is absolute value) but then i got 83~~~....then i checked the answer and it was 97~~ and i saw if i subtract 180 form 83 you get the answer..but why??? or am i just doing it wrong???

then i need to find what time the boats are closest to each other

HELP!!!

Edited May 18, 2011 by hellokitty818

[jess1ca]

You probably already figured out the solution... since this is from a while ago

I think you just forgot a negative sign (if you switch the signs the answer will differ by 180*)

EDIT: Don't absolute value it!

but I'll post it in case anyone else is reading and confused...

Boat A: x(t)=3-t, y(t)=2t-4

Direction vector: (-1,2)

Boat B: x(t)=4-3t, y(t)=3-2t

Direction vector (-3,-2)

(-1,2) DOT (-3,-2) = √[(-1)^2+(2)^2] √[(-3)^2+(-2)^2] cosX

3-4 = √5√13 cos X

-1/√5√13= cos X

X= 97.1 degrees

EDIT:

I didn't see the part about when the boats are the closest.

I'm not sure if there's another method of doing it (I'm not taking HL math) but...

The boats are the closest when their squared distance is the smallest

let Q be the squared distance...

Q = [(3-t) - (4 - 3t)]^2 + [(2t-4)-(3 - 2t)]^2

= (2t -1)^2 + (4t-7)^2

= 20t^2 - 60t + 50

Q is at a minumum when dQ/dt=0

So you differentiate:

dQ/dt = 40t - 60

0= 40t - 60

60=40t

t=1.5

Therefore the boats are the closest when t=1.5 (at 1.5 hours)

Edited May 29, 2011 by SmilingAtLife:)

[L'estrange98]

Can someone explain the same problems without using differentiation? I haven't arrived to that topic yet.

[Slovakov]

The way I'd do this without differentiation is in major part similar to what jess1ca wrote.

You have to find (or it doesn't matter). The result, similar to what is above, will be . There is a quadratic function nder the root, and here is where you would have to do the differentiation... but you dont. Because we know from the function that it has a minimum, and we know that the vertex of a quadratic function of form is at . Here a = 20 and b = -60, so it's clear that .

Now I'm aware that it's probably not the way you're required to solve this question. Since it deals with vectors, it can be solved using vector operations only, but it's a tricky process. I'm a horrible teacher and don't even dare to explain it fully, I'll just say that the trick is once you;ve found line , you need its dot product with its direction vector to equal 0 (i.e. find when these two are perpendicular). That is . The result of this equation will be your answer.

I'll get back to it when I have some more sleep.

Edited April 29, 2015 by Slovakov