(Challenge question) Submarine hunting. Two surface ships on manoeuvresare trying to determine a submarine’s course and speed to prepare for an aircraftintercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is locatedat (0, 5, 0). All coordinates are given in thousands of meters. Ship A locates thesubmarine in the direction of the vector 2i + 3j – (1/3) k, and ship B locates it in thedirection of the vector 18i – 6j – k. Four minutes ago, the submarine was locatedat (2, -1, -1/3). The aircraft is due in 20 minutes. If the submarine moves in astraight line at a constant speed, to what position should the surface ships directthe aircraft to drop the torpedo? (Hint: Write equations of lines in vector form).
Question
(Challenge question) Submarine hunting. Two surface ships on manoeuvresare trying to determine a submarine’s course and speed to prepare for an aircraftintercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is locatedat (0, 5, 0). All coordinates are given in thousands of meters. Ship A locates thesubmarine in the direction of the vector 2i + 3j – (1/3) k, and ship B locates it in thedirection of the vector 18i – 6j – k. Four minutes ago, the submarine was locatedat (2, -1, -1/3). The aircraft is due in 20 minutes. If the submarine moves in astraight line at a constant speed, to what position should the surface ships directthe aircraft to drop the torpedo? (Hint: Write equations of lines in vector form).
Solution
To solve this problem, we need to find the equation of the line that represents the submarine's path. We know that the submarine was at position (2, -1, -1/3) four minutes ago and it's moving in a straight line at a constant speed.
The equation of a line in 3D space is given by:
r = a + tb
where:
- r is the position vector of a point on the line,
- a is the position vector of a known point on the line,
- b is the direction vector of the line, and
- t is the parameter which varies.
We know that the submarine was at position (2, -1, -1/3) four minutes ago, so a = (2, -1, -1/3).
The direction vector b can be found by taking the cross product of the direction vectors given by the two ships. The direction vector given by ship A is 2i + 3j - (1/3)k = (2, 3, -1/3) and the direction vector given by ship B is 18i - 6j - k = (18, -6, -1).
So, b = (2, 3, -1/3) x (18, -6, -1) = (-3, -6, -60).
Therefore, the equation of the line representing the submarine's path is:
r = (2, -1, -1/3) + t(-3, -6, -60)
Now, we need to find the position of the submarine in 20 minutes. Since the submarine was at position (2, -1, -1/3) four minutes ago, t = 24 minutes.
Substituting t = 24 into the equation of the line gives:
r = (2, -1, -1/3) + 24(-3, -6, -60) = (-70, -145, -1441/3)
So, the surface ships should direct the aircraft to drop the torpedo at position (-70, -145, -1441/3) in thousands of meters.
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