I think you mean the line given by the vector equation

which could also be written as the parametric equations , , and with as parameter.

This line is given by or the parametric equations , , andL2:={(2) + mu(2) :mu ER}

(2) (1)

(1) (2)

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 . Setting the y coordinates equal, . That second equation clearly says that putting that into the first equation so that . But "most" lines, in three dimensions do not intersect at all. We need to check that these values also satisfy but it is obviously true that any will satisfy that. Putting into , , and gives x= 1- 1= 0, y= 2- 1= 1, and z= 1+ 2(-1)= -1.how come is (0) ?

(1)

(-1)

So , , and .-----------------------------------

ii] Find intersection of two planes

P1:={(2) + lamda1(2) +lamda2 (1) :lamda1,lamda2 ER}

(2) (1) (2)

(3) (2) (2)

Here, , , andP2:={(3) + mu1(2) +mu2 (3) :mu1,mu2 ER}

(1) (3) (2)

(3) (2) (3)

Again, on the line of intersection, , , and . 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 inhow come is {(1) + lamda1 (2) :lamda1 ER}??(0) (1)

(1) (2)fourunknown 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 : or . Putting that into the first equation we have or .

Now put those values for and into, say the third equation: or which reduces to .

We can replace with in the parametric equations for the second plane to get x, y, and z given in terms of the single parameter , 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.

W1 consists of the vectors of the form such that 5x+ 2y- z= 0. From that equation, z= 5x+ 2y so that .-------------------------------------------

iii]V=R^3 and subspaces

W1 = {(x) ER^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.

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

You are given vectors that span W2. If they are independent they willbea basis. To see if they are independent, look at

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 . In particular, we can write

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 for some numbers a and b. Any vector in W2 can be written as .

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.

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Once again thank you so much for your time and attention,

It would help alot if i understand the questions