Vectors

[APK911]

Hi,
I have a couple of questions...

1) How would you use vector methods to prove that in the parallelogram OABC, the line drawn from O to the mid-point of AB cuts AC at the point of trisection of AC that is nearer to A.

2) A defensive missile battery launches a ground to air missile A to intercept an incoming enemy missile B. At the moment of A's launch the position vectors of A and B (in meters), relative to the command HQ were:
r_A= <600, 0, 0> & r_B= <2200, 4000, 600>
A and B maintain the velocities (m/s): v_A= <-196, 213, 18> & v_B= <-240, 100, 0>
Prove that A will not intercept B and find "how much it misses by".

Suppose instead that the computer on missile A detects that its off target and, 20 sec into its flight, A changes its velocity and interception occurs after a further 15 sec. Find the constant velocity that A must maintain during this final 15 sec for interception to occur.

Sorry that it's really long... Thank you in advance!

[geminii]

I'm not sure of the answers, I'm just posting so I'll get notified when someone replies. I'd love to know the answers to these questions too!

[APK911]

I just tried question 2 again, and I got the right answer for the first part, but I'm not sure if I did it the right way, or if it was just pure luck. This is what I did.
2) <600-192t₁, 213t₁, 18t₁> = <2200-240t₂, 4000+100t₂, 600>
Solving simultaneously, there are no solutions, therefore, they do not intersect.
600-196t+213t+18t=2200-240t+4000+100t+600
t=35.43
Sub. that into A and B: A <-6344, 7546.29, 637.71> & B <-6302.86, 7542.86, 600>
A-B (BA)= <-41.14, 3.43, 37.71>
|BA|= sqrt[(-41.14)^2 + (3.43)^2+(37.71)^2]=55.9m

[APK911]

So I got some help and figured out the last part of question 2...

To find the constant velocity that A must maintain, first you'd have to find the position vectors of both A and B when t=20.
A: <-3320, 4260, 360> & B: <-2600, 6000, 600>
When interception occurs, < -3320+15x, 4260+15y, 360+15z > = < -6200, 7500, 600 > [x, y & z being A's new velocity vector]
Solving simultaneously, x= -192, y= 216 & z=16
Therefore, A must maintain a constant velocity of -192i+216j+16k

I'm still stumped by the 1st question though....

[RuiAce]

Hopefully I didn't use too many concepts not taught in the WACE course. I haven't done vector proofs in ages either.



Explanation: The equation of a line is dependent on a 'direction vector' through any two points in question. The scalars (lambda and mu) are there to ensure we have coverage over the entire line. OM passes through the origin, so this scaling is all that's necessary, but we need to specify a 'position vector' for AC to start. Here, A has been chosen.

Note that c-a is the vector from A to C.

[APK911]

Thank you very much!!