Find the equation of a plane containing two given points and having a given distance to a third point

Question

This problem is part of examination preparation material for second mid-semester test of 12-th grade in my school:

In the 3D space Oxyz, given 3 points $A(1,0,0)$, $B(0,-2,3)$, $C(1,1,1)$. Let $(P)$ be the plane containing $A$, $B$ such that the distance from $C$ to the plane $(P)$ is $\frac{2}{\sqrt{3}}$. The equation of the plane $(P)$ is:

• A. $2x + 3y + z - 1 = 0$ or $3x + y + 7z + 6 = 0$
• B. $x + y + z - 1 = 0$ or $-2x +37y+17z+13=0$
• C. $x + y +2z - 1 = 0$ or $-2x +3y+7z+23=0$
• D. $x + y + z - 1 = 0$ or $-23x+37y+17z+23=0$

This is a multiple choice question. However, we're still expected to provide some work...

It's not quite important, as it is not part of the mandatory homework section. But I still find it quite interesting, somehow.

So far, the best thing I've got is in Geogebra using some translation and rotations.

As on paper, I don't know where to even start... Surely I can't just tell my teacher "so we rotate this segment $sin^{-1}{(\frac{2}{\sqrt{3}}} \div |Vector(A,D)|)$ where D is the image of C on plane from A and normal vector AB". Also, I don't know how Geogebra's Rotate(Object, Angle, PointOfOrigin, Axis) work.

Any suggestion to an alternative approach to this problem would be appreciated...

Answer 1

Let the equation of the plane be $a x + b y + c z = d $

Since $A$ and $B$ lie on the plane, then

$ a = d $

$ -2 b + 3 c = d $

Further the distance of the plane from $C$ is $\dfrac{2}{\sqrt{3}} $. Therefore,

$ \left( \dfrac{2}{\sqrt{3}} \right)^2 = \dfrac{ (a + b + c - d)^2} {a^2 + b^2 + c^2} $

Solving the first two equations, gives

$ a = d = 2t $ , where $t \in \mathbb{R} $

$ c = 2s $, where $ s \in \mathbb{R} $

$ b = 3 s - t $

Substituting this into the third equation,

$ \dfrac{4}{3} = \dfrac{ (5 s - t)^2 }{ 4 t^2 + (3 s - t)^2 + 4 s^2 } $

so that,

$ 16 t^2 + 4 ( 9 s^2 - 6 s t + t^2 ) + 16 s^2 = 3 ( 25 s^2 + t^2 - 10 t s ) $

And finally,

$ 17 t^2 - 23 s^2 + 6 t s = 0 $

Take $ s = 1 $, then

$ 17 t^2 + 6 t - 23 = 0 $

Factor,

$(17 t + 23) (t - 1) = 0$

whose roots are

$ t =1 $ and $ t = - \dfrac{23}{17} $

For $ t = 1 $ we get

$ a = d = 2 , c = 2 , b = 2 $

So that the equation is $ x + y + z = 1 $

And for $ t = -\dfrac{23}{17}$, modify $s$ to $17$ , then $t = -23 $, and we will have,

$ a = d = -46 , c = 34 , b = 3(17) + 23 = 74 $

Dividing through by $2$, the equation is

$ - 23 x + 37 y + 17 z = -23 $

Answer 2

Finding a line in $R^{3}$ through A and B is simple enough. The problem that then arises is that since point C is not on the plane, it is not simply a case of finding two vectors in the plane etc.

I would suggest elimination of wrong answers is the approach intended by the way the question is written.

Option A: the plane $2x+3y+z-1=0$ does not go through point A so we can eliminate this option.

Option B: both planes contain points A and B and so we need to check the distance from each to the point C. The plane $-2x+37y+17z+13=0$ is not the required distance from C (the first plane in option B is the required distance though).

Option C: The plane $-2x+3y+7z+23=0$ is not the required distance from C.

So by elimination, it must be D.

Checking that points A and B are in both planes and both are the required distance from C is not too difficult to show.

Answer 3

If you want to do it by rotation of the normal vector of the plane, then first find the vector

$ V = B - A = (0, -2, 3) - (1, 0,0) = (-1, -2, 3) $

Create two orthogonal unit vectors to $V$, these are arbitrary, but an easy choice is

$ U_1 = \dfrac{1}{\sqrt{5}} (2, -1, 0) $

$U_2 = \dfrac{ V \times U_1}{\| V \times U_1 \| }$

Now $ V \times U_1 = \dfrac{1}{\sqrt{5}} ( 3 , 6 , 5 ) $

So that

$ U_2 = \dfrac{ (3, 6, 5)}{ \sqrt{70}} $

Now the equation of the plane is a function of one parameter only $\theta$

The equation is

$ N \cdot ( r - A ) = 0 $

where $N = \cos \theta \ U_1 + \sin \theta \ U_2 $

Since $N$ is a unit vector, the distance between $C$ and the plane as given in this equation is $| N \cdot (C - A) | $. And we have $C - A = (0, 1, 1) $

Therefore, we require that

$ | \cos \theta (U_1 \cdot (C - A) ) + \sin \theta (U_2 \cdot (C - A) | = \dfrac{2}{\sqrt{3}} $

Now $U_1 \cdot (C - A) = - \dfrac{1}{\sqrt{5}} $

and $U_2 \cdot (C - A) = \dfrac{11}{\sqrt{70}} $

We already know that there are only two solutions, so we can drop the absolute value operation. Hence, the equation becomes

$ - \dfrac{1}{\sqrt{5}} \cos \theta + \dfrac{11}{\sqrt{70}} \sin \theta = \dfrac{2}{\sqrt{3}} $

The solutions of this trigonometric equation in radians are

$ \theta_1 = 1.30963916 $

and

$ \theta_2 = 2.487700856 $

The corresponding normal vector $N$ is

With $\theta_1$:

$ N = \langle \dfrac{1}{\sqrt{3}} , \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} \rangle $

and $ d = N \cdot A = \dfrac{1}{\sqrt{3}} $

So that, after scaling by $\sqrt{3}$ the equation becomes:

$ x + y + z = 1 $

With $\theta_2$, we get:

$ N = \langle -0.491816896 , 0.791183702, 0.363516836 \rangle $

and $ d = N \cdot A = -0.491816896 $

So that after scaling, we get

$ -23 x + 37 y + 17 z = -23 $