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Let $P_1$ be the plane through the origin containing the vectors $[1,2,-1]$ and $[0,1,1]$. Let $P_2$ be the plane through the point $(1,1,1)$ parallel to the vectors $[-1,2,2]$ and $[3,4,-2]$

I know how to find the standard form of a plane that passes through a point and contains a line, but not one that contains two lines.

Find an equation for $P_1$ in normal form $ax+by+cz=0$

The planes $P_1$ and $P_2$ intersect in a line. Find a parametric equation for this line.

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Please don't just dump exercises here. Tell us what you know about the problem, what you've been able to do, where you have gotten stuck, why you want to know how to do the problem, and so on, and so forth. –  Gerry Myerson Oct 11 '12 at 3:29
    
this looks like a homework problem to me. no one should answer it. –  karen Oct 11 '12 at 3:40
    
@karen, a lot of the problems posted to this website are homework problems, and that is accepted here. Some of us are happy to give hints about homework problems, and not give complete solutions. There has been much discussion of this at the meta site. –  Gerry Myerson Oct 12 '12 at 12:16
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1 Answer

up vote 1 down vote accepted

A related problem. If you know the normal vector $n=(n_1,n_2,n_3)$ to a plane and a point $p=(x_0,y_0,z_0)$ lies in the plane, then we can find the equation of the plane as $$ n.(X-p)=0 \,,$$ where $X=(x,y,z)$ an arbitrary point lies in the plane. The point is not a problem, since you have three of them $p_1=(0,0,0)\,,p_2=(1,2,-1)\,, p_3=(0,1,1)$. The task is how to find the normal vector to the plane. I believe, you have studied the cross product of two vectors and you know the fact that the cross product of two vectors is a vector perpendicular to the plane that contains these two vectors.

Now, since you have three points, you can form two vectors

$$ v_1=p_2-p_1 \,, \quad v_2 = p_3-p_1 \,.$$

Once you form $v_1$ and $v_2$ you can find the normal to the plane as

$$ n = v_1 \times v_2 \,.$$

Now, you should be able to find the equation of the plane $P_1\,.$

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