Question

A yacht is sailing at a constant speed of 510 km-1 in the direction –i-3j. Initially it is at point (-6,10). A beacon is at (0,0) at the centre of a tiny atoll. (c.) If there is a reef of radius 8 km around the atoll, will the yacht hit the reef? Answer: shortest dist=8.85 km, so it will miss the reef. *This is a vector problem * I have no idea what to do. Something to do with perpendicular distance and dot product maybe? Any help appreciated =]

Answer 1

So if we draw it:


I drew two circles since we don't know if it is in or out.

Answer 2

Here is the vector from the starting point.
Actually it will work for any vector on our line.

Answer 3

You can get the angle between these vectors by using the dot product: \[
|a\cdot b| = |a||b|\cos \theta\implies \theta = \cos^{-1}\left(\frac{|a\cdot b|}{|a||b|}\right)
\]

Answer 4

Suppose \(a\) was the vector for out path and \(b\) was the vector from our initial point to the center.
The then the smallest distance from us to the center is: \[
|b|\sin \theta
\]

Answer 5

Remember that: \[
\cos^2\theta +\sin^2\theta = 1\implies |\sin\theta| = \sqrt{1-\cos^2\theta }
\]

Answer 6

We expect \(\sin \theta\) to be positive so \(|\sin \theta|=\sin \theta\).

Answer 7

This means we get: \[
|b|\sqrt{1-\cos(\theta)} = |b|\sqrt{1-\cos^2\left(\cos^{-1}\left(\frac{|a\cdot b|}{|a||b|}\right)\right)} = |b|\sqrt{1-\left(\frac{|a\cdot b|}{|a||b|}\right)^2}
\]

Answer 8

Might as well pull the \(|b|\) in:\[
\sqrt{|b|^2-\frac{(a\cdot b)^2}{|a|^2}}
\]

Answer 9

Also \[
|x|=\sqrt{x\cdot x}
\]So \[
\sqrt{b\cdot b-\frac{(a\cdot b)^2}{a\cdot a}}
\]

Answer 10

@karenli324 ez

Answer 11

Oh, it was a mistake to put absolute value bars around \(a\cdot b\), by the way. I dropped the ball on that. Just pretend they aren't there.