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Here is my vector:


And my plane: Created from the following vectors:

$x: (3,0,1)+t(-1,1,2)$

$x: (0,2,-1)+s(2,-2,-4)$


(Cartesian: $x+y-3=0$)

I've tried to do the following:

vector: $(-3+4r, 1, -4+r)$

plane: $(3-t+2n, t-2n, 1+2t-4n)$


1) $3-t+2n = -3+4r$

2) $t-2n = 1$

3) $1+2t-4n = -4+r$

But I didn't find the right solution...
The intersection point should be: $(1,1,-3)$(written in the math book)

What did I do wrong? Thanks in advance.

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up vote 1 down vote accepted

A rather straightforward way to solve this is following these 2 steps:

1) find the cartesian equation of the plane, i.e. its equation of the form $ax+by+cz+d=0$. In order to do this, just remember that the vector $(a,b,c)$ is perpendicular to the plane and observe that you can easily obtain two independent vectors $\vec v$ and $\vec w$ parallel to the plane (so that for instance $(a,b,c)=\vec v\wedge\vec w$).

2) impose that the generic point of the line satisfies the equation of the plane and solve for $t$.

I leave the details to you as an exercise.

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I've found b (by doing cross product between: (2,-2,-4) and (-1,1,2), b=-1,x=z=0) how can I find "d"? – funerr Jan 25 '13 at 18:04
Once you have the coefficients $a$, $b$, $c$, the last coefficient $d$ can be found just imposing that any point of the plane (e.g. $(3,0,1)$) satisfies the equation. – Andrea Mori Jan 25 '13 at 18:10

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