Submarine coordinates + speed

[jovial sand]

Hi I need some help here finding out the boats position when it reaches the surface.

Translation:

A submarine is observed at time t = 0 at the point (2, 1, −1/3) and 20 minutes later at (1, 3, −1/4).
Here the water surface lies in the xy-plane, the z-axis points vertically upwards and the length is nautical miles (nm).
Now assume that the submarine keeps a straight course with constant speed.
i. What is the submarine's speed, v, and velocity, v = |v| (measured in knots = NM/h) ? I got this to, 6.7128 NM/h = 12,432 km/h. Correct?

ii. Where and when does the submarine reach the surface? This one I dont know, my try:

Delta r = r1 - r0 = (2, 1, -1/3) - (1, 3, -1/4) = (-3, 6, 1/4)

Then we can find the velocity vector by dividing the change in position by the time:
V = Delta r / (1/3) = (3, -6, -1/4) / (1/3) = (9, -18, -3/4) NM/h.

[grizzled stirrup]

(Just FYI, the symbol for nautical mile is NM or nmi, to avoid confusing it with nanometers.)

[lofty flare]

I. The speed should be a vector: v = {-3, 6, 1/4}. The velocity is |v| = √(721)/4, which is approximately equal to what you wrote.
By the way, for a second, I thought nm meant nanometers, as it usually does, and was a bit confused 😄
II. You know the direction vector of a line and a point that it goes through (even two points).
So, write its equation and find the value of parameter t such that r(t) = {something, something, 0}.

[lofty flare]

Ah, I see, that makes sense.

[jovial sand]

changed it!

[jovial sand]

sorry for the units confusion

[lofty flare]

No worries, I instantly felt something was up 😄
It was quite funny, though.

[grizzled stirrup]

It wasn't really confusing because we weren't dealing with both units in this problem, it's just a general rule of thumb.

"M" is also an acceptable symbol for nautical miles, but I didn't mention it because that one might actually be confusing.

[jovial sand]

wait a minute, (-3, 6, 1/4) how?

[lofty flare]

Difference of positions divided by time, as usual.

[jovial sand]

Yeah ik but I got this, (3, -6, -1/4)

:S

[lofty flare]

You mixed up the initial and final positions.

[jovial sand]

oh yeah I see it lol, i did x1-x2 instead

[jovial sand]

@lofty flare I think it should be like this {-3, 6, 1/4} + t(2, 1, -1/3) ?

or am I completely off lol

[lofty flare]

No. The initial point should be the first point and direction vectors is the velocity.

[jovial sand]

So this, (2, 1, -1/3) + t{-3, 6, 1/4}??? Should I get the dot product from these two points?

[lofty flare]

Yes, but what do you want to use the dot product for?

[jovial sand]

I dont honestly know, been watcchin too many dot product videos I guess lol.

So, r = (2-3t, 1+6t, -1/3+1/4t) ?

[lofty flare]

Yes. And you need to find the time at which the z-coordinate becomes zero.

[jovial sand]

okay so this equation (2-3t, 1+6t, -1/3+1/4t) = 0

[lofty flare]

No.

The other coordinates can be whatever.

We need to surface, so only the z-coordinate needs to become zero.

[jovial sand]

-1/3+1/4t = 0 so t = 4/3

(-2, 9, 0)?

oo so 4/3h to reach the surface and coordinates should be '(-2, 9, 0) ?

[lofty flare]

Yes.

[jovial sand]

Thanks for your help btw!

[sterile parrot]

@jovial sand has given 1 rep to @lofty flare