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Topic: perpendicular vectors..URgeNT (Read 1985 times) |
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Deepesh Baburajan
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Find all vectors that are perpendicular to both u1 = (1,1,4,3) and
u2 = (1,0,1,3)
b) verify that the set W of all the vectors 'w'found in (a) is a subspace of R4
i would have thought that the first part,there are infinite vectors that are perpendicular...
basically, to find it... i would find u1.b and then u2.b wher b is
(a,b,c,d)
u1.b = (1a+1b+4c+3d) and
u2.b = (1a+1c+3d)
the i would have to equations
1a + 1b + 4c + d = 0
1a + 1c + 3d = 0
from here, i would use iteration to find the solutions... But the above equation means that there are infintely many solutions... is there a GENERAL equation that i could find perhaps? and wat does part b ask exactLY?
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towr
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Re: perpendicular vectors..URgeNT
« Reply #1 on: May 3rd, 2004, 4:37am » |
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I think b) asks to show that you don't simply get incidental points in vectorspace, but a subspace (for instance a hypercube, or hypersphere of some other shape). Since the vectors are 4 dimensional, and you're looking for a perpendicular vector to just two of them you can within certain limits vary the coordinates in a continuous way and still have perpendicular vectors.
I suppose you can first find a vector v perpendicular to u1 and u2, and then a vector w perpendicular to v,u1 and u2. Any linear combination of v and w is then perpendicular to u1 and u2 as well (so effectively you have a 2D plane in [bbr]4 full of perpendicular vectors)
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| « Last Edit: May 3rd, 2004, 4:51am by towr » |
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Icarus
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Re: perpendicular vectors..URgeNT
« Reply #2 on: May 4th, 2004, 6:25pm » |
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Part 1 is asking you to find the SET of all solutions, and yes, the set is infinite. It is, as towr indicates, a plane in [bbr]4. Being a plane, your answer will need to look like this:
{(Ar+Bs, Cr+Ds, Er+Fs, Gr+Hs) | r, s [in] [bbr]} for some constants A, B, C, D, E, F, G, H. Every value of r & s will provide you with a vector orthogonal to your original 2.
Part B is asking you to prove that this set (call it W) is a subspace of [bbr]4. If you do not know what a subspace is by now, I am very curious what course it is you are taking? Surely they have defined this term before giving you this problem?
A subset of [bbr]n is a "subspace" if it is closed under addition and scalar multiplication: For all x, y [in] W and c [in] [bbr], x + y and cx are both in W. When this holds, W is a vector space itself. What you need to show in (b) is that these two things hold true.
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Icarus
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Re: perpendicular vectors..URgeNT
« Reply #3 on: May 4th, 2004, 6:29pm » |
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on May 3rd, 2004, 4:37am, towr wrote:| I think b) asks to show that you don't simply get incidental points in vectorspace, but a subspace (for instance a hypercube, or hypersphere of some other shape). |
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No. By subspace, they are refering to a sub vector space. Not a hypercube or hypersphere. It will be a line or plane or 3-space that passes through the origin. But how you show it is exactly what I said in the other post: show that the solution set is closed under addition and scalar multiplication.
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Deepesh Baburajan
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Re: perpendicular vectors..URgeNT
« Reply #4 on: May 4th, 2004, 8:02pm » |
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on May 4th, 2004, 6:25pm, Icarus wrote:Part 1 is asking you to find the SET of all solutions, and yes, the set is infinite. It is, as towr indicates, a plane in [bbr]4. Being a plane, your answer will need to look like this:
{(Ar+Bs, Cr+Ds, Er+Fs, Gr+Hs) | r, s [in] [bbr]} for some constants A, B, C, D, E, F, G, H. Every value of r & s will provide you with a vector orthogonal to your original 2.
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ok, so it is infinite as i thought... But how do i find Ar+Bs, Cr+Ds, Er+Fs, Gr+Hs) how do i find these values?
was my previous approach of dot product,folowd by solving a system of equations correct? this is an extension question, for bonus marks... So i've had to learn most of this myself using text books etc. i had to learn the dot product, and subspaces etc.
so if you could kindly explain it in a layman manner, it owuld be greatly appreciated...
thanku
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towr
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Re: perpendicular vectors..URgeNT
« Reply #5 on: May 5th, 2004, 1:34am » |
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You're pretty much on the right way
on May 3rd, 2004, 3:46am, Deepesh Baburajan wrote:1a + 1b + 4c + 3d = 0
1a + 1c + 3d = 0 |
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1a + 1b + 4c + 3d - (1a + 1c + 3d) = 0 - 0
1b + 3c = 0
b=-3c
1a + 1c + 3d = 0
a = -c - 3d
so we get:
(-c-3d, -3c, c, d) = c*(-1, -3, 1, 0) + d*(-3, 0, 0, 1)
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| « Last Edit: May 5th, 2004, 1:48am by towr » |
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Deepesh baburajan
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Re: perpendicular vectors..URgeNT
« Reply #6 on: May 5th, 2004, 5:47am » |
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thanx, but i think i need more help on part 2... how exactly do i prove that it's a subspace? and watever else i need to prove?
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towr
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Re: perpendicular vectors..URgeNT
« Reply #7 on: May 5th, 2004, 6:25am » |
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Icarus already said how you can prove it's a subspace, it must be closed under addition and multiplication by a scalar
We have
W = {c*(-1, -3, 1, 0) + d*(-3, 0, 0, 1) | c [in] [bbr] [wedge] d [in] [bbr]}
And what you now have to prove is
[forall]u,v [in] W: u+v [in] W
and
[forall]u [in] W,s [in] [bbr]: s*u [in] W
which should follow easily enough from the description of W
(note: u [in] W [equiv] [exists]c,d[in][bbr]: u=c*(-1, -3, 1, 0) + d*(-3, 0, 0, 1) )
The only thing I'm not sure about is the 0-vector, personally I wouldn't want it in W as it's not really perpendicular to anything, but for closure under addition it should be in W.
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| « Last Edit: May 5th, 2004, 6:26am by towr » |
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Deepesh baburajan
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Re: perpendicular vectors..URgeNT
« Reply #8 on: May 5th, 2004, 6:57am » |
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sorry,wat do those weird symbols mean? i know i have to prove that it's a subspace, it must be closed under addition and multiplication by a scalar...but i don't get HOW i prove it,from the defintion of W
as i mentioned before, this is an extension,so i haven't learnt it
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towr
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Re: perpendicular vectors..URgeNT
« Reply #9 on: May 5th, 2004, 8:01am » |
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[bbr] is the set of real numbers
[in] means 'in', or 'is an element of'
i.e. a [in] [bbr] means a is an element from the set of real numbers
S = { expression | condition } means
S is the set of all values of the expression for which the condition holds true
i.e C(r) = { (a,b) | a2+b2=r2 ] is the set C(r) of all points on a cirkel with radius r
( C(1) would be the set of all points on a cirkel with radius 1, C(2) the points with radius 2 etc )
If it's not specified it's generally assumed the numbers (in this case a,b and r) are from [bbr]
[exists] means 'exists'
i.e. [exists]a[in][bbr]: a > 0 means there exists a number a in the set of real number for which it is true that a is greater than 0
[forall] means 'for all'
i.e. [forall]a[in][bbr]: [exists]b[in][bbr]: b > a means for all real numbers a there exists a real number b which is larger than a.
[bbr] is a good example of a set that is closed under addition and multiplication.
The addition of any two real numbers is itself a real number, and the multiplication of two real numbers is also always a real number.
The thing you need for your proof is the fact that any vector in W is determined uniquely by the two real numbers c and d. And then you can use the fact that [bbr] is a vector space.
Alternatively, you can show that there's a linear transformation (for example using a matrix) from S={(c,d) | c,d[in][bbr]} = [bbr]2 to W. Since S is a vector space W must be one as well (because for any linear transformation T, T(a+b) = T(a) + T(b) and T(s*a) = s*T(a) ).
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| « Last Edit: May 5th, 2004, 8:27am by towr » |
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Deepesh Baburajan
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Re: perpendicular vectors..URgeNT
« Reply #10 on: May 6th, 2004, 7:01am » |
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thanku guys soo much for ur helP!
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