Two boats are moving along straight paths and their position vectors at noon are:

For the first question, i assumed its initial position would be 0 hours after noon. So i just answered it as:

[tex]7\mathbf{i}-5\mathbf{j}[/tex]

Now i wouldnt have a clue if thats even close to being correct, but its the only decent answer i could come up with.

For the next part, i drew the diagram, and just found a vector going from the position of the first boat to the second, from the initial positions, which gave me the final vector:

[tex]4\mathbf{i}+18\mathbf{j}[/tex]

For part c, the velocity of the second boat, i just wrote how much the position vector increases for every incriment of t. I came up with:

[tex]-3\mathbf{i}-\mathbf{j}[/tex]

The problem with this was that the question said they were travelling in a straight line, and if i apply this velocity, then they change their direction. So im lost.

And with the final question, im stuck, because i cant really do it until i answer the previous ones correctly.

So if anyone doesn't mind, please put me on the right track for these questions, because i highly boubt ive answered them correctly.

could you explain how you got 5i + 13j please? I re did it, and ended with the i component being 5, but i dont understand how you got 13 for the j component.

Anyway, im doing the final question now. So to find when they collide, i need to find then their position vectors are the same, at the same value for t.

So do i just equate the components for each vector?: i 7-4t j -5+t

i 12-3t j 13-t

So then i need to find when the i and j components of both ships are the same. I got that when the i component of both boats is 27, the j component will be 4, so they will collide at:

You can check if your answer is correct (whether they collide at your defined point), choose a vector equation and sub your values in for i and j such as this;

[tex]r_{1} = (7-4t)j + (t-5)j[/tex]

Subbing [itex]27i + 5j[/itex] into each component;