1. ## Vectors

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:
rA= <600, 0, 0> & rB= <2200, 4000, 600>
A and B maintain the velocities (m/s): vA= <-196, 213, 18> & vB= <-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!

2. ## Re: Vectors

2). $r_A(t)=\left<600-196t,213t,18t \right>$

$r_B(t)=\left<2200-240t,4000+100t,600\right>$

If the missiles collide, they have to be at the same position at the same time ... the only time this may occur is when $18t=600 \implies t=\dfrac{100}{3} \, sec$. Are both of the other postion components equal to each other at that time?

Distance between the missiles at any time $t$ Is ...

$D=\sqrt{[r_{Ax}(t)-r_{Bx}(t)]^2 + [r_{Ay}(t)-r_{By}(t)]^2 + [r_{Az}(t)-r_{Bz}(t)]^2}$

minimizing the distance involves a derivative or the use of a graphing utility.

For the course change of missile A ...

1. Determine both missile positions at $t=20$. Recommend re-starting the clock to zero at this position for both missiles to sync things up.

2. Determine new position vectors as functions of time for both missiles using a variable for missile A's new velocity. Set each component equal for new $t=15$ and solve for missile A's velocity components.

3. ## Re: Vectors

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

4. ## Re: Vectors

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

**Sorry I think I accidentally posted this twice**