Find equation of a plane that passes through point P $(-1,4,2)$ that contains the intersection line of the planes $$\begin{align*} 4x-y+z-2&=0\\ 2x+y-2z-3&=0 \end{align*}$$

Attempt: I found the the direction vector of the intersection line by taking the cross product of vectors normal to the known planes. I got $\langle 1,10,6\rangle$. Now, I need to find a vector normal to the plane I am looking for. To do that I need one more point on that plane. So how do I proceed?

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    $\begingroup$ The direction vector alone does not suffice: You need to find a point that lies on the intersection line. Of course, this point also lies in the plane you are looking for. $\endgroup$ – m_l Apr 27 '12 at 11:37

Put the two given equations together in a system, set $x=0$ (say) and solve for $y,z$ to get an actual point on the intersection line. Accompanied by its direction vector and the outlying point $P$, we can determine three points from the desired plane and subsequently determine an equation for it.

Dostre's edit (see comments)

$x=0:\begin{cases} y-2z-3=0 \\ -y+z-2=0 \end{cases}$ ;$\;\;\;\;$ add them up and you get:

$\;\;\;\;\;\;\;\;\;\;\begin{cases} -z-5,\;\;z=-5\\ -y+z-2=0,\;y=-7 \end{cases}\Rightarrow$Point $ (0,-7,-5)$, call it Q, is on the line of intersection.

So now we have two points $P(-1,4,2)$ and $Q(0,-7,-5)$ on our desired plane and vector w thats is parallel to the desired plane. In order to find the equation of the desired plane we need a vector that is normal to it.We can find that normal vector by taking cross product of two vectors that are parallel to the desired plane. We already have w so the other vector will be


Now normal vector to desired plane will be the cross product of w and PQ:

PQ x w=$\begin{vmatrix} i & j & k \\ 1 & -11 & -7 \\ 1 & 10 & 6 \end{vmatrix}=i\begin{vmatrix} -11 & -7 \\ 10 & 6 \end{vmatrix}-j\begin{vmatrix} 1 & -7 \\ 1 & 6 \end{vmatrix}+k\begin{vmatrix} 1 & -11\\ 1 & 10 \end{vmatrix}=4i-13j+21k$

$4i-13j+21k=<4,-13,21>$ is the vector normal to the desired plane.

  • $\begingroup$ I see yes. I was trying that method. Three different times I set one of the parameters to zero and solved the system of equations. I get a point say Q that lies on the intersection line. So now I have two points P and Q that I know are lying in the looked for plane. So I get a vector PQ and take the cross product of PQ and the direction vector of the line of intersection. I did not look for the third point since the direction vector of the intersection line is parallel to the looked for plane. However, I do not get the answer that is in the book. $\endgroup$ – Koba Apr 27 '12 at 11:51
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    $\begingroup$ @Dostre: When I set $y=0$, I got the point $L=(0,-7,-5)$. I then computed the displacement $\vec{LP}=(-1,11,7)$. Crossing with the intersection line's direction vector, $$(1,10,6)\times(-1,11,7)=(4,-12,21)$$ as a normal vector to our desired plane. Do these match with your computations? What is the book's answer? $\endgroup$ – anon Apr 27 '12 at 12:01
  • $\begingroup$ you mean you set $x=0$. Your answer is almost correct. You made little mistake calculating the y point in the normal vector to our desired plane. It is $<4,-13,21>$. Yep. Everything worked out. I guess every time I was making an arithmetic mistake. Thanks. You was helpful. May I edit your answer by adding an explicit way of gettin to the normal vector of the desired plane, so other people can see it clearly? $\endgroup$ – Koba Apr 27 '12 at 13:29
  • $\begingroup$ @Dostre: Ah, of course it's -13, and yes I meant $x=0$. :-) Go ahead and edit, I don't mind. $\endgroup$ – anon Apr 27 '12 at 21:38

Consider the family of planes $u(4x-y+z-2)+(1-u)(2x+y-2z-3)=0$ where $u$ is a parameter. You can find the appropriate value of $u$ by substituting in the coordinates of the given point and solving for $u$; the value thus obtained can be substituted in the equation for the family to yield the particular member you need.

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    $\begingroup$ or as special case $(4x-y+z-2)-(2x+y-2z-3)=0$ if the previous equation has no solution, because $(1,-1)$ can never be in the form $a*(u, 1-u)$. $\endgroup$ – carlop Apr 27 '12 at 12:28
  • $\begingroup$ thanks. We did not cover that in class $\endgroup$ – Koba Apr 27 '12 at 13:48
  • $\begingroup$ An even general approach (that don't have the problem of addressing special case) is calling $a=(4x-y+z-2), b=(2x+y-2z-3)$ (evaluated in the point $P(-1,4,2)$, so $a=4*(-1)-(4)+(2)-2=-8, b=2*(-1)+(4)-2*(2)-3=-5$), and the solution is $b*(2x+y-2z-3)-a*(4x-y+z-2)=0$, that should be $22x-13y-6z-1=0$. If both $a$ and $b$ evaluate to $0$ means that the point is on the intersection line of the two planes, so the problem is indeterminated. $\endgroup$ – carlop Apr 27 '12 at 13:49
  • $\begingroup$ @carlo, maybe that should be an answer instead of a comment. :) $\endgroup$ – J. M. is a poor mathematician Apr 27 '12 at 13:51
  • $\begingroup$ The answer actually is $4x-13y+21z=-14$. Does not look like these planes are the same. $\endgroup$ – Koba Apr 27 '12 at 17:36

Find any point that belongs to the line by looking for the intersections of this line with the coordinate planes. E.g., put $z=0$ and find $x$ and $y$ from the system $$ 4x-y=2,\quad 2x+4=3. $$

  • $\begingroup$ Thanks Artem. Now I got the idea $\endgroup$ – Koba Apr 27 '12 at 13:48

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