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## 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;

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

The second line;

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

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

And the equation of the plane;

[tex]A(x-x_0) + B(y-y_0) + C(z-z_0)=0[/tex]

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

[tex]-(x-2) + 2(y-1) + (z-2)=0[/tex]

[tex]-x + 2y + z - 2=0[/tex]

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;

[tex]-3 + 0y + 3 - 2 = -2[/tex]

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