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astarmathsandphysics:

--- Quote from: avrila on February 24, 2009, 10:55:03 am ---Can you please help me with this:

1. Find the fourth vertex of the parallelogram:
A(2,1,0) B(3,2,-1) C(0,-1,1)

2. Find the shortest distance between point A (3,5,6) and line (x-7)/2=y-5=(z-12)/6

Thanks in advance...

--- End quote ---

AD=BC=(0,-1,1)-(3,2,-1)=(-3,-3,2) so D=A+(-3,-3,2)=(-1,-2,2)

2.eequation of line=(7,5,12)+t(2,1,6). Suppose they intersect at (a,b,c) then (a-3,b-5,c-6) is at right angles to (2,1,6) but (a,b,c) is on line so (a,b,c)= (7+2t,5+t,12+6t)
 and (4+2t,t,6+6t).(2,1,6)=0 ie  44+41t=o so t=-44/41 sthe sub into  (7+2t,5+t,12+6t) to find a and use pythagoras theorem in 3D to find distance between A and (a,b,c)

avrila:
Next question..

How would i solve this system of equations by hand of course. Is it possible?
625A -125B +25C -5D +E =4.5
112A -34.3B +10.6C -3.25D+E=2.25
0.00391A -0.0156B +0.0625C -0.25D +E=6.25
92.4A + 29.8B +9.61C + 3.10 D + E= 1.40
625A +125B +25C +5D +E =3.85

Help would be very much appreciated. Thanks.

astarmathsandphysics:
My impression is that for IB you would expected to use your calculators solve function. You are doing ib?

avrila:
Ya.. I'm doing IB. The question is not from IB though. Someone ask me and I'm just curious to answer. I try them but it seems little bit complex. Just wondering if you could solve the problem.. There must be a way right. Someone said use augmented matrices. I usually do this but up to three only not till 5x5 matrices.

Well, I'll keep on trying. Do tell me if you manage to solve this.. Thanks.

avrila:
What is arbitrary point in vector?
A question ask, Find the position vector of an arbitrary point R which lies on the line.

I just want to know what is arbitrary point at the moment. ..

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