Find an equation of the plane that contains these lines.?

Question

line 1 passes through (1,1,0) and parallel to the vector v=<1,-1,2> and line 2 passes through (2,0,2) and parallel to w=<-1,1,0>.

Answer 1

The equation of the plane is:
x + y = 2

There are several ways to solve this kind of problem. I'll show you the way that I did it.

I like to start by finding 3 points on the plane. Then, I use the 3 points to find the equation of the plane. Of course, the three points must not be collinear. (They must not all lie along one straight line.)

We have two points given in the question. We know that (1,1,0) and (2,0,2) are on the plane. So we only need to find one other point.

There is something a little bit tricky to be aware of here. The point (2,0,2), which is in the definintion of line 2, happens to be on line 1. This means that we cannot choose the third point to be on line 1, because that would make the three points collinear. Therefore, we must choose a point that is on line 2, but NOT on line 1.

Let's pick a new point that is on line 2. To do that, we take some constant multiple of the vector w, and add that to the point (2,0,2). Let the constant multiplier be 1.

1 * <-1,1,0> + (2,0,2) = (1,1,2)

Therefore, our three points are:
(1,1,0)
(2,0,2)
(1,1,2)

In general, the equation of a plane looks like:
A x + B y + C z = 1
So, we need to know the values of A, B, and C that will make the equation hold for all three of our chosen points.

This is a problem of linear algebra. Form a matrix equation as follows:

[1,1,0] [A] = [1]
[2,0,2] [B] = [1]
[1,1,2] [C] = [1]

The solution can be found by a matrix inversion and multiplication.
A = 1/2
B = 1/2
C = 0

So, the equation of the plane is:
(1/2)x + (1/2)y + (0)z = 1

We can simplify that to:
(1/2)x + (1/2)y = 1

And then simplify further to:
x + y = 2

Answer 2

Find an equation of the plane that contains these lines.

L1(s) = P + sv = <1, 1, 0> + s<1, -1, 2>
L2(t) = Q + tw = <2, 0, 2> + t<-1, 1, 0>
where s and t are scalars ranging over the real numbers

First we need to find if the lines are coplanar. If they are they will be either parallel or they will intersect. A moment's inspection will show that they intersect at the point Q(2, 0, 2). This point is on L1 at s = 1.

The normal vector n, to the plane containing the lines will be orthogonal to both the directional vectors, v and w, of the plane. Take the cross product.

n = v X w = <1, -1, 2> X <-1, 1, 0> = <-2, -2, 0>

Any non-zero multiple of n is also a normal vector to the plane. Divide by -2.

n = <1, 1, 0>

With the normal vector n and the a point on the plane we can write the equation of the plane containing the two lines. Let's choose point Q(2, 0, 2).

1(x - 2) + 1(y - 0) + 0(z - 2) = 0
x - 2 + y + 0 = 0
x + y - 2 = 0

Answer 3

I guess answer of the book is correct, so u have 2 do that math again in another way...