I have come across a difficult problem in my textbook.

The problem ask:

Find an equation of the plane that passes through the line of intersection of the planes $x-z=1$ and $y+2z=3$ and is perpendicular to the plane $x+y-2z=1$.

So far I found the direction vector of the line of intersection to be $<1,−2,1>$ and I have identified a point on this line when $x=0$ to be $(0,5,−1)$.

I do not know how to find the desired plane from here.

Any assistance would be appreciated.


The plane should be parallel to the vector $a=<1,-2,1>$ and $b=<1,1,-2>$. Now we need a vector which is normal to these two vectors as a result a normal to the plane. Let $c=a \times b$ where $\times$ represent cross product. By definition of cross product, $c$ is normal to the plane. So, let $c$ be $<m,n,o>$. The equation of the plane is $$mx+n(y-5)+o(z+1)=0$$.

  • $\begingroup$ What's your comment for? $\endgroup$ – Mhenni Benghorbal Jan 18 '13 at 5:48
  • $\begingroup$ I think the normal to the plane should be normal to two vecrors. <-1,2,1> and <1,1,-2>, since the plane contains the line x−z=1 and y+2z=3. Or am I missing something here? $\endgroup$ – Mohan Jan 18 '13 at 5:52
  • $\begingroup$ How did you get the vector <-1,2,-1>? $\endgroup$ – jascal Jan 18 '13 at 14:45
  • $\begingroup$ Sorry, it should be <1,-2,1> . The direction vector of the line. $\endgroup$ – Mohan Jan 18 '13 at 14:53
  • $\begingroup$ Your plane should be parallel to two vectors <1,-2,1>, the direction vector of the line and <1,1,-2>, normal of the plane x+y-2z=1. $\endgroup$ – Mohan Jan 18 '13 at 15:09

First, you found the direction vector of the line, which is the intersection of the planes $x-z=1$ and $y+2z=3$, correctly: $$ \vec{l}=(1,-2,1). $$ You can take your point as $$ M=(0,5,-1). $$ You are also required that your plane is perpendicular to $x+y-2z=1$, which means that the normal to your sought plane $\vec{n}$ has to be orthogonal to $\vec{n}_1=(1,1,-2)$ (normal of the plane). So, you are looking for the plane passing through the point $M$ and such that its normal $\vec{n}\perp \vec{l}$ and $\vec{n}\perp \vec{n}_1$. To find such $\vec{n}$ you need to use the cross product $\vec{n}=\vec{l}\times\vec{n}_1$, or, in coordinates: $$ \vec{n}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{vmatrix}=3\mathbf{i}+3\mathbf{j}+3\mathbf{k}, $$ hence your vector is $\vec{n}=(1,1,1)$.

Finally, you find your plane: $$ (x-0)+(y-5)+(z+1)=0, $$ or $$ x+y+z=4. $$


A plane in Euclidean space is determined completely by the following:

1) A direction vector perpendicular to the plane, and

2) A single point on the plane.

You say you've found a point that is in the plane you wish to describe. Can you get a perpendicular vector? (Hint: look at the second condition.) And once you have 1) and 2), do you know how to write the equation of the plane determined by them?

  • $\begingroup$ why isn't the direction vector of the line of intersection ? The plane should be parallel to it. $\endgroup$ – Mohan Jan 18 '13 at 5:36
  • $\begingroup$ @Mohan A 3-dimensional plane has infinitely many directions that are parallel to it, so a single parallel direction vector tells you little about the orientation of the plane in space. (If you had 2 parallel and independent vectors you'd be in good shape.) If you know what the direction perpendicular to the plane, though, then this tells you how the plane is oriented in space. $\endgroup$ – Gyu Eun Lee Jan 18 '13 at 5:38
  • $\begingroup$ It is also perpendicular to the plane x+y-2z=1. So, I think we can get two parallel directions <-1,2,1,> and <1,1,-2>. $\endgroup$ – Mohan Jan 18 '13 at 5:41
  • $\begingroup$ Ah yes. My mistake - for some reason I thought the perpendicular plane was enough. $\endgroup$ – Gyu Eun Lee Jan 18 '13 at 5:44

A related problem. Already, you have got a point which lies in the plane. What's left is the normal to the plane. You are given that the plane is perpendicular to a plane which its normal is $n_1=(1,1,-2)$. To find the normal to the plane, assume

$$ n=(a,b,c) \implies n.n_1=0 \implies (a,b,c).(1,1,-2)=0\implies a+b-2c=0. $$

Solve for $a,b$, and $c$ to get the normal vector. You will have infinite number of solutions, just pick up one. One possible solution is $n=(1,1,1)$.

  • $\begingroup$ The plane should contain the intersection of x-z=1 and y+2z=3. $\endgroup$ – Mohan Jan 18 '13 at 5:32
  • $\begingroup$ @Mohan: Already he has a got a point which lies in the plane. You can find the normal to the plane since it is perpendicular to another plane. $\endgroup$ – Mhenni Benghorbal Jan 18 '13 at 5:54
  • $\begingroup$ But the plane should also be parallel to <1,-2,1>? $\endgroup$ – Mohan Jan 18 '13 at 5:57

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