Author Topic: Maths/Physics help  (Read 23633 times)

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #45 on: February 24, 2009, 12:19:05 pm »
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...

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)

Offline avrila

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Re: Maths/Physics help
« Reply #46 on: February 27, 2009, 10:19:21 pm »
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.
IB May 2009

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #47 on: February 27, 2009, 10:24:27 pm »
My impression is that for IB you would expected to use your calculators solve function. You are doing ib?

Offline avrila

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Re: Maths/Physics help
« Reply #48 on: February 27, 2009, 11:11:12 pm »
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.
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Offline avrila

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Re: Maths/Physics help
« Reply #49 on: February 28, 2009, 06:17:12 am »
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. ..
IB May 2009

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #50 on: February 28, 2009, 08:25:26 am »
Yes use augmented matrices. Row reduce and the inverse is on the right hand side. The multiple the vector on right hand side of the original system of equations by the inverse you have found and the answer is the vector  you have found

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #51 on: February 28, 2009, 08:35:02 am »
Yes use augmented matrices. Row reduce and the inverse is on the right hand side. The multiple the vector on right hand side of the original system of equations by the inverse you have found and the answer is the vector  you have found
suppose the equation of a line is r(t) =2i+4j-k+t(5i-2j+7k) the arbitrary point vector is (2+5t,4-2t,-1+7t)

Offline avrila

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Re: Maths/Physics help
« Reply #52 on: March 01, 2009, 01:05:33 am »
P=(4,1,-1) Q=(3,3,5) R=(1,0,2c) S=(1,1,2)

Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.
IB May 2009

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #53 on: March 01, 2009, 07:54:44 am »
P=(4,1,-1) Q=(3,3,5) R=(1,0,2c) S=(1,1,2)

Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.
QR=(-2,-3,2c-5) and PR=(-3,-1,2c+1)
QR.PR=6+3+(2c-5)(2c+1)=9+4c^2-8c-5=4c^2-8c+4=0 solution is c=1

Offline avrila

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Re: Maths/Physics help
« Reply #54 on: March 01, 2009, 02:08:23 pm »
Ya.. I know where my mistake is. I misplace one of the sign and ruin the question.. Thanks.
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Re: Maths/Physics help
« Reply #55 on: March 02, 2009, 02:43:57 am »
Plane pi has equation r.(2i +j + 2k)= 36
Point B and C has position vectors (i +3 +2k) and (3i +2j + k) respectively.

>Find the position vector of the point where the line OB meets pi and find the perpendicular distance from B to pi.
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #56 on: March 02, 2009, 11:33:31 pm »
Plane pi has equation r.(2i +j + 2k)= 36
Point B and C has position vectors (i +3 +2k) and (3i +2j + k) respectively.

>Find the position vector of the point where the line OB meets pi and find the perpendicular distance from B to pi.
OB=t(1,3,2) so subsitute into pi x=t, y=3t, z=2t

We have (t,3t,2t).(2,1,2)=36 so 9t=36 and t=4 and the line meets pi at(4,12,8). The distance is \surd(3^2+9^2+6^2)=3\surd14
« Last Edit: March 03, 2009, 07:50:00 am by astarmathsandphysics »

Offline avrila

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Re: Maths/Physics help
« Reply #57 on: March 03, 2009, 04:30:40 am »
Ok.I got it until (4,12,8). And this is the coordinates where OB meets the plane. In other words, it is the point of intersection between the line and the plane.

But to find the distance from B to pi, why we just use the same coordinate? B has it owns coordinate right?...
Can you please explain.
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #58 on: March 03, 2009, 07:50:48 am »
I corrected my post

Offline avrila

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Re: Maths/Physics help
« Reply #59 on: March 03, 2009, 10:02:40 am »
Ok.. I got the answer but it is slightly different.
For the distance, instead of 3,6,9 I got 6,3,6. After substitute this in the distance formula, I got 9 as the answer.

Hope I am in the right track.
IB May 2009