Question

The pairs of vectors v(1)=(-1,1,0) and v(2)=(1,0,1) span a plane P in R(3). The pairs of vectors w(1)=(0,1,0) and w(2)=(1,1,0) span a plane Q in R(3).

Show that P and Q are different and compute the subspace of R(3) that is given by the intersection P(intersect)Q?

@helder_edwin

Answer 1

If you have some time to help.

Answer 2

To show that they are diferent you can think like this: if they were equal then the vector product of one of the vectors of one plane by one of the vectors of the other plane, must give a vector that is perpendicular to both planes, and this vector must be paralel to the vector that is defined by the formula of the plane, therefore their vector product must be zero. When you do it, you'll see that it is not zero and then the planes are not the same.

Answer 3

Or you can find the vector that is given by the formula by doing the vector product of the 2 vectors they already give you.

Answer 4

Did I manage to explain it? I believe its kind of confusing.

Answer 5

okay, so if i do the cross product of v1 and v2, then do the cross product of w1 and w2, are you saying that they should be the same or multiplies of each other and that would mean they would be the same plane, and if the cross products are different then they would be different planes? is that what you are saying?

Answer 6

Yes thats it, if the cross products are multiples of each other then they are the same, and vice versa

Answer 7

okay... well that solves the first part... now how do you find subspace intersection P(intersect)Q

Answer 8

You first need to find a point that belongs to both planes, then you take the same cross products, and make their cross products to find a vector thatbelongs to both planes, with a point and a vector, you have the line.

Answer 9

and how do you find that point?

Answer 10

@TuringTest do you know how to find this intersection

Answer 11

uggh i need help rissa lol @Smile4rissa

Answer 12

I took the cross product of the first pair and got (1,1,-1) and the cross product of the second pair and got (0,0,-1) so that means that since they are different vectors then P and Q are different planes? But I don't understand how to find the intersection?

Answer 13

i found an example online that took the sum of the two vectors in the span of each and set them equal to each other.
so P=span{(1,-1,0), (0,1,1)} .... (x,-x+y,y)
Q=span{(1,0,0),(0,1,0)} .... (a,b,0)
so setting those two equal
x=a
-x+y=b
y=0

but then i got lost what the example did after this... but I am not sure if this is even right

Answer 14

give a second. i remember there's a very easy way: i'm looking for it.

Answer 15

ok

Answer 16

r u familiar with strang's books on linear algebra?
if so, this is in one if them.

Answer 17

no i am not

Answer 18

first find the null space of
\[ \large \begin{pmatrix}
-1 & 1 &0 & 1\\ 1 & 0 & 1 & 1\\ 0 & 1 & 0 & 0
\end{pmatrix} \]

Answer 19

u should read them (just as reference) they're very good.

Answer 20

I found another example that took the null space of P-Q. but I wasn't sure about that either.. but I will do that one quick

Answer 21

okay so I got (-1,0,-2,1)

Answer 22

i got the null space is spanned by (1,0,-2,1)

Answer 23

ohh yes you are right, sign error on my part.. so is that the intersection?

Answer 24

no.

Answer 25

so then once we find the null space, how do we use that to find the intersection?

Answer 26

the first three columns have pivots. and the fourth column is the independent variable.
right?

Answer 27

\[ \large \color{red}{1}\cdot(-1,1,0)+\color{red}{0}\cdot(1,0,1)
\color{red}{-2}\cdot(0,1,0)+\color{red}{1}\cdot(1,1,0)=(0,0,0) \]
agree?

Answer 28

yes I do

Answer 29

how do I use that to find the intersection?

Answer 30

the first three vectors correspond to the pivots and the last one to the free variable.

Answer 31

so u can write
\[ \large 1(-1,1,0)-2(0,1,0)=-1(1,1,0) \]
and this vector spans the intersection.

Answer 32

\[ \large P\cap Q=\langle(1,1,0)\rangle \]

Answer 33

okay so it is just a line then.

Answer 34

yes

Answer 35

was that correct above, to take the cross products of each set to see if the planes are different?

Answer 36

well. what u proved with that is that they are not parallel. hence they must intersect.

also, remember that P and Q aren't just planes, they r subspaces. so, should u have obtained parallel normal vectors then that would have been a proof that P and Q are equal.

This would not work with planes (translated subspaces of dim = 2), though.

Answer 37

is there a better way to show that P and Q are different?

Answer 38

i don't know if it is better, but you could prove that the vectors that span P don't belong to Q and viceversa.

but the hint @ivanmlerner gave you is pretty cool.

Answer 39

would it matter which vector you took from which plane when doing that?

Answer 40

@ivanmlerner
what you said did kind of confuse me.

Answer 41

What I sugested was:
v1xv2 gives a vector that is perpendicular to P
w1xw2 gives a vector that is perpendicular to Q
Since all vectors that belongs to P are perpendicular to the first cross product and all the vectors that belong to Q are perpendicular to the second cross product. Because of this when you do the cross product of the cross product: (v1xv2)x(w1xw2) you must get a vector that belongs to both planes. Now, this vector gives a direction to the line that is the intersection of both planes, and all you need now is a point to locate this line in space that belongs to both planes.
Then, the formula for the line is: p+ (v1xv2)x(w1xw2) where p is the point.

Answer 42

Oh, wait, they dont give you any point in P or Q, so they are general planes. Then you cant find the point p but the direction is still ok.

Answer 43

would I still get (1,1,0) doing it this way?

Answer 44

Let me see

Answer 45

b/c I think I got (-1,1,0)

Answer 46

I got this too

Answer 47

so was what I did earlier to get (1,1,0) wrong?
doing it this way that should be the intersection right?

Answer 48

Look, I didn't have linar algebre so I don't understand the method above, but I think this one is correct, but it might not, @helder_edwin could you take a look here?

Answer 49

i don't think that \((v_1\times v_2)\times(w_1\times w_2)\) will necesarily span the intersection of P and Q.

Answer 50

strangs book you mentioned earlier is the same guy that teaches at the MIT?

Answer 51

I found somewhere that said this
\[P(x)=Q(y) so P(x)-Q(y)=0\]
and then the matrix would be
\[\left[\begin{matrix}-1 & 1 & 0 & -1\\ 1 & 0 & -1 & -1\\ 0 & 1 & 0 & 0\end{matrix}\right]\]
and then the Null space=(-1,0,-2,1)
but I don't know if this makes sense or not.

Answer 52

u can easily prove that (1,1,0) is both in P and Q.
the following systems MUST be consistent:
\[ \large \left(\begin{array}{cc|c}
-1 & 1 & 1\\ 1 & 0 & 1\\ 0 & 1 & 0
\end{array}\right)\qquad\qquad
\left(\begin{array}{cc|c}
0 & 1 & 1\\ 1 & 1 & 1\\ 0 & 0 & 0
\end{array}\right) \]

Answer 53

yes, @ivanmlerner . Strang teaches at MIT.

Answer 54

that makes sense.

Answer 55

Then why is my answer wrong? I just want to know.

Answer 56

yes, it is pretty much the same. since u r looking for the intersection of two spaces for which u have bases then
\[ \large av_1+bv_2=cw_1+dw_2 \]
is the system whose solutions (if they exist) are in both spaces.

Answer 57

Sorry I have a very limited knowledge of linear algebra, my answer was based on geometry.
Could you try to explain to me wy my answer is wrong with geometric arguments?

Answer 58

i think u r right.

Answer 59

the first system is inconsistent

Answer 60

wait. you think his answer is right?

Answer 61

actually it is.

Answer 62

i'm checking what went wrong with my approach.

Answer 63

sorry. i just can't figure out what went wrong. what i did should have given us the base of the intersection.

Answer 64

the problem with @ivanmlerner 's geometric approach is that it only works in R^3. so u need a more general method.

Answer 65

Yes, that's true.

Answer 66

@zonazoo . if u leave this post open, i'll try to post tomorrow night the answer using linear algebra.

Answer 67

I would, but we can still respond if the post is closed right?.. I have two more problems that I am really having trouble with that I want to post.

Answer 68

i definitely still want to find if there is a more general approach

Answer 69

@helder_edwin is it not possible to use my mthod with more dimensions by analogy?

Answer 70

yes. it would make sense. but in dimensions higher than 3, we have nothing ressembling the cross product.