Author Topic: Vectors  (Read 3628 times)

Offline Light

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Vectors
« on: February 01, 2010, 08:32:28 am »
1)the line T has vector equation r=2i+s(i+3j+4k)
find the position vectorS of the points on T which are exactly square root of 250 units from origin.
determine the position vector of point on T which is closest to point with position vector 6i-j+3k.

2)the centre line of an underground railway tunnel follows a line given by R=x(10,8,-1)for 0 or equal<t< or equal 40 the units being metres.the centre line of another tunnel at present stops at point wit position vector (200,100,-25) and it is proposed to extend this in a direction (5,7,Z).the constant Z has to be chosen so that at the point where one tunnel passes over the other,there is at least 15 metres difference in depth between centre lines of two tunnels.what restrictions does this impose on value Z?another requirement is that the tunnel must not be inclined at more than 5 degree to horizontal.What values of z satisfy both?

Offline astarmathsandphysics

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Re: Vectors
« Reply #1 on: February 01, 2010, 09:46:55 am »
When I get home

Offline astarmathsandphysics

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Re: Vectors
« Reply #2 on: February 01, 2010, 10:25:43 am »
1)the line T has vector equation r=2i+s(i+3j+4k)
find the position vectorS of the points on T which are exactly square root of 250 units from origin.
determine the position vector of point on T which is closest to point with position vector 6i-j+3k.
250=(2+s)^2+(3s)^2+(4s)^2
250=4+4s+s^2+9s^2+16s^2=4+4s+25s^2
25s^2+4s--246=0
s=-3.218 or 3.058
r=2i+(-3.218 or 3.058)(i+3j+4k)


d^2=|(2i+s(i+3j+4k)-(6i-j+3k.)|^2=(-4+s)^2+(3s+1)^2+(4s-3)^2
=16-8s+s^2+9s^2+6s+1+16s^2-24s+9=26s^2-26s+26
differentiate 52s-26=0 so s=1/2
then d^2=26*1/2^2-26*1/2+26=19.5 so d=sqrt(19.5)

Offline astarmathsandphysics

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Re: Vectors
« Reply #3 on: February 01, 2010, 10:41:05 am »
R=x(10,8,-1)

explain this?

Offline Light

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Re: Vectors
« Reply #4 on: February 02, 2010, 10:54:09 am »
the ans for ques 1 is 5i+9j+12k.and -1/13(15i+123j+164k).no so many decimal.

d^2=|(2i+s(i+3j+4k)-(6i-j+3k.)|^2=(-4+s)^2+(3s+1)^2+(4s-3)^2
=16-8s+s^2+9s^2+6s+1+16s^2-24s+9=26s^2-26s+26
differentiate 52s-26=0 so s=1/2
then d^2=26*1/2^2-26*1/2+26=19.5 so d=sqrt(19.5)
i dont get this part.mind explaning.the position vector(ans) is 5/2 i +3/2 j+2k.whats the working?

sorry for the R=x(10,8,-1).cant type the vector in matrix form.its actually line r= X(10i,8j,-1k) where X is the parameter.
thanx for solving.

Offline astarmathsandphysics

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Re: Vectors
« Reply #5 on: February 02, 2010, 10:58:54 am »
use s or t for parameters. I will try it now

Offline astarmathsandphysics

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Re: Vectors
« Reply #6 on: February 02, 2010, 11:15:21 am »
I have looked at this question.
r_1=t(10,8,-1)
r_2=(200,100,-25)+s(5,7,Z)=(200+5s,100+7s-25+sZ)
d=|r_1-r_2|=sqrt((200+5s-10t)^2+(100+7s-8t)^2+(25+sZ+t)^2)
expand and simplify then differentiate with respect to s,t,Z and solve three simultaneous equations. Not a typical A level question. What module is it?

Offline Light

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Re: Vectors
« Reply #7 on: February 02, 2010, 11:27:48 am »
it is. from Cie A level pure maths 3 and 4 textbook.the miscellaneous ex.i dont get the diffrentiation part.what does it have to do with vectors?the earlier ques u mind explain also?dont get it.
« Last Edit: February 02, 2010, 11:30:58 am by Light »

Offline astarmathsandphysics

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Re: Vectors
« Reply #8 on: February 02, 2010, 11:30:46 am »
No clear line between degreen and A level I think. Have you enough to do it or shall I work it through? If I do I whink it will be on paper

Offline Light

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Re: Vectors
« Reply #9 on: February 02, 2010, 11:32:11 am »
it would be nice if u did that.put it in adobe reader.