# Where is the mistake in a point closest to cross section of two planes and a line problem

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?

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

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

Hint.

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?

• Yes I solved it. I understand.
– 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

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