Finding First Line & Plane for Points (2,1,2), (3,0,3), (1,-1,2) & (0,1,3)

[Yann]

Homework Statement

I must make a first line with the points (2,1,2) and (3,0,3)

Then another line with the points (1,-1,2) and (0,1,3)

Then, I must find the equation for a plane including the first line and perpendicular to the second line. It should be very easy...

2. The attempt at a solution

The first line;

\mathbf{r}(t) = (2+t)\mathbf{i} + (1-t)\mathbf{j} + (2+t)\mathbf{k}

The second line;

\mathbf{R}(t) = (1-t)\mathbf{i} + (-1+2t)\mathbf{j} + (2+t)\mathbf{k}

The normal vector is \mathbf{N} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}

And the equation of the plane;

A(x-x_0) + B(y-y_0) + C(z-z_0)=0

Where (A,B,C) are taken from the Normal (-1,2,1). I take the point (2,1,2) on the second line for (x_0,y_0,z_0), so I get the final equaiton;

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

-x + 2y + z - 2=0

The problem is; it doesn't work. If I take the point (3,0,3) on the first line, it is out of the plane;

-3 + 0y + 3 - 2 = -2

BTW, I don't want the answer, I just want to know what I'm doing wrong.

 

[HallsofIvy]

The equations for your lines are correct but your normal vector is not.

A vector in the direction direction of the first line is obviously
\mathbf{i} - \mathbf{j} + \mathbf{k}

but that does not have dot product with \mathbf{N} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k} equal to 0.

 

[Yann]

But that's what I don't understand, by definition, the normal is the nonzero vector perpendicular to the plane. If my equation for the second line is ok, then the vector N = -i+j+2k should be perpendicular to the plane (because I must find a plane perpendicular to the second line), but it's impossible. If I choose another vector for the normal, the plane can't be perpendicular to the second line (by definition).

There must be something very simple I don't get, but I really can't see it.

Thx for the your help

 

[arunbg]

Since yourworking seems fine to me I would say that such a plane is impossible.
The normal specifies the orientation of the plane and for it to contain the point (2,1,2), it becomes uniquely determined. Whether another given point lies in it is a different matter that should be checked as you have done.

 

[HallsofIvy]

Homework Statement

I must make a first line with the points (2,1,2) and (3,0,3)

Then another line with the points (1,-1,2) and (0,1,3)

Then, I must find the equation for a plane including the first line and perpendicular to the second line. It should be very easy...

2. The attempt at a solution

The first line;

\mathbf{r}(t) = (2+t)\mathbf{i} + (1-t)\mathbf{j} + (2+t)\mathbf{k}

The second line;

\mathbf{R}(t) = (1-t)\mathbf{i} + (-1+2t)\mathbf{j} + (2+t)\mathbf{k}

The normal vector is \mathbf{N} = -\mathbf{i} + 2\mathbf{j} + \mathbf{k}

Okay, I misunderstood. This is the normal to the plane, and so in the direction of the line, not normal to the line which is what I thought you were saying.

And the equation of the plane;

A(x-x_0) + B(y-y_0) + C(z-z_0)=0

Where (A,B,C) are taken from the Normal (-1,2,1). I take the point (2,1,2) on the second line for (x_0,y_0,z_0), so I get the final equaiton;

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

-x + 2y + z - 2=0

The problem is; it doesn't work. If I take the point (3,0,3) on the first line, it is out of the plane;

-3 + 0y + 3 - 2 = -2

BTW, I don't want the answer, I just want to know what I'm doing wrong.

The fact that the plane contains one point of the first line does not necessarily mean that it contains the entire line!
The real problem is that, if the second line is normal to the plane, then it is normal to every line in it. Here, the given second line is NOT normal to the given first line because the dot product of the vectors point along the lines is not 0. Far from being "very easy", this is impossible. Since the two given lines are not perpendicular, there is no plane, containing the first line, normal to the second.

 

[Yann]

It was an error by my teacher. The point (1,-1,2) should be (-1,-1,2), and I solved this easily :)