# Intersection of lines,planes,vector spaces

• Aug 12th 2010, 08:21 PM
jht
Intersection of lines,planes,vector spaces
hi i got some question that i dont understand,hope someone could help me with them,thanks so much.:'(

i] Find intersection of two lines

L1:={(1) + altha(1) :altha E R }
(2) (1)
(1) (2)

L2:={(2) + mu(2) :mu E R }
(2) (1)
(1) (2)

how come is (0) ?
(1)
(-1)

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ii] Find intersection of two planes

P1:={(2) + lamda1(2) +lamda2 (1) :lamda1,lamda2 E R }
(2) (1) (2)
(3) (2) (2)

P2:={(3) + mu1(2) +mu2 (3) :mu1,mu2 E R }
(1) (3) (2)
(3) (2) (3)

how come is {(1) + lamda1 (2) :lamda1 E R}??
(0) (1)
(1) (2)

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iii]V=R^3 and subspaces

W1 = {(x) E R^3 :5x +2y -z =0 }
(y)
(z)
W2=span:{(1) , (2) , (4) }
(1) (1) (3)
(1) (3) (5)
find bases for W1,W2,W1+W2 and intersection of W1 and W2.

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Once again thank you so much for your time and attention,
It would help alot if i understand the questions
• Aug 13th 2010, 04:32 AM
HallsofIvy
Quote:

Originally Posted by jht
hi i got some question that i dont understand,hope someone could help me with them,thanks so much.:'(

i] Find intersection of two lines

L1:={(1) + altha(1) :altha E R }
(2) (1)
(1) (2)

I think you mean the line given by the vector equation $\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix} + \alpha\begin{bmatrix}1 \\ 1 \\2 \end{bmatrix}$
which could also be written as the parametric equations $x= 1+ \alpha$, $y= 2+ \alpha$, and $z= 1+ 2\alpha$ with $\alpha$ as parameter.

Quote:

L2:={(2) + mu(2) :mu E R }
(2) (1)
(1) (2)
This line is given by $\begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix}+ \mu\begin{bmatrix}2 \\ 1\\ 2\end{bmatrix}$ or the parametric equations $x= 2+ 2\mu$, $y= 2+ \mu$, and $z= 1+ 2\mu$

Quote:

how come is (0) ?
(1)
(-1)
Of course, at the point of intersection, the x, y, and z values (or corresponding components of the vector) must be equal. Setting the x coordinates equal, we have $x= 1+ \alpha= 2+ 2\mu$. Setting the y coordinates equal, $y= 2+ \alpha= 2+ \mu$. That second equation clearly says that $\mu= \alpha$ putting that into the first equation $1+ \alpha= 2+ 2\alpha$ so that $\mu= \alpha= -1$. But "most" lines, in three dimensions do not intersect at all. We need to check that these values also satisfy $z= 1+ 2\alpha= z= 1+ 2\mu$ but it is obviously true that any $\alpha= \mu$ will satisfy that. Putting $\alpha= -1$ into $x= 1+ \alpha$, $y= 2+ \alpha$, and $z= 1+ 2\alpha$ gives x= 1- 1= 0, y= 2- 1= 1, and z= 1+ 2(-1)= -1.

Quote:

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ii] Find intersection of two planes

P1:={(2) + lamda1(2) +lamda2 (1) :lamda1,lamda2 E R }
(2) (1) (2)
(3) (2) (2)
So $x= 2+ 2\lambda_1+ \lambda_2$, $y= 2+ \lambda_1+ 2\lambda_2$, and $z= 3+ 2\lambda_1+ 2\lambda_2$.

Quote:

P2:={(3) + mu1(2) +mu2 (3) :mu1,mu2 E R }
(1) (3) (2)
(3) (2) (3)
Here, $x= 3+ 2\mu_1+ 3\mu_2$, $y= 1+ 3\mu_1+ 2\mu_2$, and $z= 3+ 2\mu_1+ 3\mu_2$

Quote:

how come is {(1) + lamda1 (2) :lamda1 E R}??
(0) (1)
(1) (2)
Again, on the line of intersection, $x= 2+ 2\lambda_1+ \lambda_2= 3+ 2\mu_1+ 3\mu_2$, $y= 2+ \lambda_1+ 2\lambda_2= 1+ 3\mu_1+ 2\mu_2$, and $z= 3+ 2\lambda_1+ 2\lambda_2= 3+ 2\mu_1+ 3\mu_2$. In the first problem, we had three equations in two unknown values which normally doesn't have a solution just a two line don't normally intersect in three dimensions. Here we have three equations in four unknown values which is "under determined"- there will be an infinite number of solutions just as two planes intersect on a line, not just at one point.

If we subtract the second equation from the third we eliminate $\lambda_2$: $1+ \lambda_1= 2- \mu_1+ \mu_2$ or $\lambda_1= 1- \mu_1+ \mu_2$. Putting that into the first equation we have $2+ 2- 2\mu_1+ 2\mu_2+ \lambda_2= 3+ 2 \mu_1+ 3\mu_2$ or $\lambda_2= -1+ 4\mu_1+ \mu_2$.

Now put those values for $\lambda_1$ and $\lambda_2$ into, say the third equation: $3+ 2(1- \mu_1+ \mu_2)+ 2(-1+ 4\mu_1+ \mu_2)= 3+ 2\mu_1+ 3\mu_2$ or $3+ 5\mu_1+ 4\mu_2= 3+ 2\mu_1+ 3\mu_2$ which reduces to $\mu_2= -3\mu_1$.

We can replace $\mu_2$ with $3\mu_2$ in the parametric equations for the second plane to get x, y, and z given in terms of the single parameter $\mu_1$, the parametric equations of a line. It might not be precisely the answer given- a single line can have an infinte number of different parametric representations depending upon how the parameter is found.

Quote:

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iii]V=R^3 and subspaces

W1 = {(x) E R^3 :5x +2y -z =0 }
(y)
(z)
W2=span:{(1) , (2) , (4) }
(1) (1) (3)
(1) (3) (5)
find bases for W1,W2,W1+W2 and intersection of W1 and W2.
W1 consists of the vectors of the form $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ such that 5x+ 2y- z= 0. From that equation, z= 5x+ 2y so that $\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ 5x+ 2y\end{bmatrix}$ $= \begin{bmatrxi}x \\ 0 \\ 5x\end{bmatrix}+ \begin{bmatrix}0 \\ y \\ 2y\end{bmatrix}$ $= x\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}$.

That should make it clear what a basis for W1 is.

You are given vectors that span W2. If they are independent they will be a basis. To see if they are independent, look at $a\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}+ b\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}+ c\begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}$ $= \begin{bmatrix}a+ 2b+ 4c \\ a+ b+ 3c \\ a+ 3b+ 5c\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

That gives the three equations a+ 2b+ 4c= 0, a+ b+ 3c= 0, and a+ 3b+ 5c= 0. If we subtract the second equation from the first, we eliminate a: b+ c= 0. If we subtract the first equation from the third, we also eliminate a: b+ c= 0 again. Those are the same equation so we can take any b and c, as long as b+ c= 0. For example, b= 1, c= -1. Putting those into the first equation, a+ 2b+ 4c= a+ 2- 4= 0 or a= 2. The given vectors are NOT independent because $2\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}+ \begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}- \begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$. In particular, we can write $\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}= \begin{bmatrix}4 \\ 3 \\ 5\end{bmatrix}- 2\begin{bmatrix}1 & 1 & 1\end{bmatrix}$

Since one vector can be written in terms of the other two, we don't need that vector. Since the two remaining vectors are not multiples of one another, they form a basis for this space. (In fact, any two of these three vectors will form a vector.)

One way to form a basis for the direct sum of the subspaces, W1+ W2, is to combine the two bases. If that set of vectors is independent, then it is a basis. (They can't be because you have four vectors and this is a subspace of dimension 3 or less.) If they are dependent, at least one of the vectors can be writen in terms of the others and you can simply discard that vector.

To find a basis for the intersection, you have to see what that intersection is: which vectors are in both spaces.

Using the bases above, any vector in W1 can be written as $a\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ b\begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}= \begin{bmatrix}a \\ b \\ 5a+ 2b\end{bmatrix}$ for some numbers a and b. Any vector in W2 can be written as $c\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}+ d\begin{bmatrix}2 \\ 1 \\ 3\end{bmatrix}= \begin{bmatrix}c+ 2d \\ c+ d \\ c+ 3d\end{bmatrix}$.

Any vector in the intersection of the two subspaces can be written in both ways: we must have a= c+ 2d, b= c+ d, and 5a+ 2d= c+ 3d.

Quote:

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Once again thank you so much for your time and attention,
It would help alot if i understand the questions
• Aug 13th 2010, 11:55 AM
jht
sorry i dont understand this part

''W1 consists of the vectors of the form such that 5x+ 2y- z= 0. From that equation, z= 5x+ 2y so that [/tex]= x\begin{bmatrix}1 \\ 0 \\ 5\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}[/tex].''
• Aug 16th 2010, 03:50 AM
HallsofIvy
Was the difficulty that my LaTex didn't work? I have corrected that.

If not, the problem is that W1 consists of all <x, y, z> such that 5x+ 2y- z= 0. From the equation, z= 5x+ 2y so <x, y, z>= <x, y, 5x+ 2y>= <x, 0, 5x>+ <0, y, 2y>= x<1, 0, 5>+ y<0, 1, 2>. That is, any vector in the set can be written as a linear combination of <1, 0, 5> and <0, 1, 2> so those span the subset. Since neither is a multiple of the other, they are also independent and so form a basis for the subspace.