The problem:

Find the point that lies on the line that is the intersection of two planes:

plane one: $$3x-2y+z = 2$$

plane two: $$x+y+2z = 4$$

The point also needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$

My try: First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.

What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$

I got: $z = 7y +2$ and $x = \frac{-5y}{3}$

Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$

So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$

Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$

Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$

So the equation of this plane is: $4z = 4*\frac{1}{2}=2$

I got $z_0$ form the last coordinate from the point.

Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.

The correct solution is: $T(1,1,1)$

Where did I go wrong?

  • $\begingroup$ $(-5/3,1,9)$ doesn't lie on plane $2$. $\endgroup$ Feb 7, 2021 at 19:02

5 Answers 5


One error is in "I got: $z = 7y +2$ and $x = \frac{-5y}{3}$".


You are doing the problem in a much more complicated way than necessary.

The second condition implies that the point lies in the plane that is perpendicular to the vector $(0,0,4)$, and contains the point $(1,-1,1)$. This plane is given by $ z=1 $.

Now you have two equations in $x$ and $y$: $$ 3x-2y=1,\quad x+y=2 $$


You have incorrectly solved the system of equations of planes since $(-5/3,1,9)$ doesn't lie on plane $2$. On solving you get $z=2-y,x=y$ so the correct choice of point lying on both planes for $y=1$ is $(1,1,1)$.

Your second error is in finding the equation of plane containing points equidistant from $A,B$: you wrote $OA+\frac12AB$ but found $OA+\frac12\color{red}{OB}$. The correct point lying on the plane is $$OA+\frac12AB=\frac{OA+OB}2=(1,-1,1)$$giving the required plane as $z=1$. Can you continue?

  • 1
    $\begingroup$ Yes I solved it. I understand. $\endgroup$
    – VLC
    Feb 7, 2021 at 19:29

It's because the midpoint between $A$ and $B$ is $(1,-1,1)$. So, the plane which consists of all the points that have the same distance from $A$ and $B$ is the plane $z=1$, not the plane $z=\frac12$.


I would have done this a slightly different way. The two planes are 3x−2y+z=2 and x+y+2z=4. From the equation of the second plane, x= 4- y- 2z. Replace x in the first equation by that: 3(4- y- 2z)- 2y+ z= 12- 3y- 6z- 2y+ z= 12- 5y- 5z= 2 so 5y+ 5z= 10 and y+ z= 2 so z= 2- y and x= 4- y- 2(2- y)= 4- y- 4+ 2y= y.

Taking y= t, parametric equations for the line of intersection are x= t, y= t, z= 2- at. Any point on that line of intersection is of the form (x, y, z)= (t, t, 2- t).

The distance, squared, from A(1, -1, -1) to (t, t, 2- t) is (t- 1)^2+ (t+ 1)^2+ (3- t)^2.

The distance, squared, from B(1, -1, 3) to (t, t, (2- t)) is (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.

For a point on the line equally distant from A an(td B, we must have (t- 1)^2+ (t+ 1)^2+ (3- t)^2= (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.

Those first two squares on the left cancel the first two on the right leaving (3- t)^2= (1+ t)^2. Taking the square root of both sides 3- t= 1+ t so 2t= 2 and t= 1. Taking 3- t= -(1+ t)= -1- t does not give a solution.

So the point on the line equidistant from A and B is (t, t, 2- t)= (1, 1, 2- 1)= (1, 1, 1).

Check: Is (1, 1, 1) on the plane 3x- 2y+ z= 2? Yes,3- 2+ 1= 2. Is (1, 1, 1) on the plane x+ 2y+ z= 4? Yes 1+ 2+ 1= 4.

The distance, squared, from (1, 1, 1) to A(1,-1,- 1) is (1- 1)^2+ (1-(-1))^2+ (1- (-1))^2= 0+ 4+ 4= 8. The distance, squared, from (1, 1, 1) to B(1, -1, 3) is (1- 1)^2+ (1- (-1))^2+ (1- 3)^2= 0+ 4+ 4= 8.

Yes, (1, 1, 1) is equally distant from A and B.


When you got $z = 7y +2$ and $x = \frac{-5y}{3}$ (which, I believe, isn't correct), that describe the line of intersection right there. Finding the cross product of the normals is needlessly complicated.

Given the points $A(1,-1,-1),B(1,-1,3)$, you can find the plane of equidistance through the methods discussed here: Equation of a plane equidistant from two points $A$ and $B$?

Taking $N = B-A=(0,0,4)$ and $(A+B)/2= (1,-1,1)$, we get $0x+0y+4z=4$, which gives $z=1$. Now you have the simultaneous equations

$$3x-2y+z = 2$$ $$x+y+2z = 4$$ $$z = 1$$


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