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### Yacht Distances and Vectors

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Date: 09/05/98 at 20:43:47
From: Zoe Stokes
Subject: Distances between lines, vectors

I have been given the following assignment:

Yacht A has initial position (-10,4) and has velocity vector [2,-1].
Yacht B has initial position (3,-13) and has velocity vector [-1,3].

In this investigation you will plot the path of each yacht and
determine the time when they are nearest and the shortest distance
they are apart.

1. Explain why the position of each yacht at time "t" is given by
r(A)=[-10,4] + t[2,-1] and
r(B)=[3,-13] + t[-1,3].
2. On squared paper plot the path of the yachts when t=O,1,2,3,4,5,...
3. Find the position vector of B relative to A.
4. Use 3. to show that if "d" is the distance between the yachts at
any time "t" then d^2 = 25t^2 - 214t + 458.
5. Show that d^2 is a minimum when t = 4.28.
6. Hence, find the time when the "d" is a minimum and then find the
shortest distance.

I have already done questions 1 and 2, but I am having problems

Thank you,
Zoe
```

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Date: 09/06/98 at 11:32:01
From: Doctor Anthony
Subject: Re: Distances between lines, vectors

Question 3:

The position of B relative to A is r(B) - r(A) = [13,-17] + t[-3,4],
which in (x,y) coordinates will be [(13-3t), (-17+4t)].

Question 4:

The formula for finding the distance apart is:

d = sqrt[(13-3t)^2 + (-17+4t)^2]

Thus: d^2 = 169 - 78t + 9t^2 + 289 - 136t + 16t^2
=  25t^2 - 214t + 458

Question 5:

To find the minimum of d^2, set the derivative equal to 0.
So 50t - 214 = 0. Thus t = 214/50 = 4.28.

Question 6:

The time when d is to be a minimum is the same time as when d^2 is a
minimum, so closest approach occurs at t = 4.28. So put t = 4.28
into the expression for d:

d = sqrt[(13-3t)^2 + (-17+4t)^2]
= sqrt[(0.16)^2 + (0.12)^2]
= sqrt(0.04)
=  0.2 miles

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus
High School Linear Algebra

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