Write the equation of a plane passing through points $A(1, 0, 1)$ and $B(0, -1, 1)$ and located at a distance of $4$ from point $C(5, 0, -3)$.

Question

Write the equation of a plane passing through points $A(1, 0, 1)$ and $B(0, -1, 1)$ and located at a distance of $4$ from point $C(5, 0, -3)$.

Attempt

What steps I should take to solve this question as after putting value of the length I am not getting anything from that. I have tried solving by taking one point M on the plane, such that MC becomes perpendicular to the plane. And using it as normal to write the equation of the plane with point and normal equation form... but I didn't work as there remains many variables.

Attempt 2

Answer 1

Alternative solution:
Let $\pi \ldots ax+by+cz+d=0$ is the required plane.
$A \in \pi \implies a + c + d = 0 \quad\qquad (1)$
$B \in \pi \implies -b+c+d=0 \,\, \qquad (2)$

$(1)-(2) \iff a + b = 0 \iff \boxed{a = -b}$.

Next, we will use now an another condition, i.e. $d(C, \pi) = 4$, or $$ \left| \frac{5a-3c+d}{\sqrt{a^2+b^2+c^2}} \right| = 4, $$ or $$ \left| \frac{4a-4c+\overbrace{a+c+d}^0}{\sqrt{a^2+a^2+c^2}} \right| = 4, $$ or $$ 4|a-c| =4\sqrt{2a^2+c^2}, $$ or $$ 16a^2+32ac=0 \iff 16a(a+2c)=0 \iff \boxed{a=0} \text{ or } \boxed{a=-2c}. $$ Finally, in the first case, i.e. for $a=0$, we have $b=0$, and form $(1)$ we can get $c=-d$.
In the second case, i.e. for $a=-2c$ from $(1)$ we can get $c=d$, and then from $(2)$ we will have $b=2d \, (=2c=-a)$.
Hence, there are two required planes:
$\pi_1 \ldots z - 1 = 0$,
$\pi_2 \ldots 2x-2y-z-1=0$.

Answer 2

The equation of a plane passing through A is

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

where we can take $n$ (the normal vector) to be a unit vector.

We now have that

$ n \cdot ( B - A ) = 0 $

and

$ | n \cdot (C - A) | = 4 $

Plugging in the numbers these two equations are

$ n \cdot ( (0,-1,1) - (1,0,1) ) = 0 $

i.e. $n \cdot (-1, -1, 0) = 0 $

And

$ | n \cdot ( (5,0,-3) - (1, 0, 1) ) | = 4 $

i.e. $ | n \cdot (4, 0, -4) | = 4 $,

Dividing by $4$ this becomes $| n \cdot (1, 0, -1) | = 1 $

Now $ n = (n_1, n_2, n_3) $ with $n_1^2 + n_2^2 + n_3^2 = 1 $

Substituting this into the two equations that we have,

$ n_1 + n_2 = 0 $ and $ | n_1 - n_3 | = 1 $

Case I: $n_1 - n_3 = 1$.

From the first equation, $n_2 = - n_1$, and from the second equation $n_3 = n_1 - 1 $

Hence,

$ n_1^2 + (- n_1)^2 + (n_1 - 1)^2 = 1 $

i.e.

$ 3 n_1^2 - 2 n_1 = 0 $

So that $ n_1 = 0 $ or $ n_1 = \dfrac{2}{3} $

Case II: $n_1 - n_3 = -1$.

In this case , $n_3 = n_1 + 1$, therefore

$ n_1^2 + (-n_1)^2 + (n_1 + 1)^2 = 1 $

So that $3 n_1^2 + 2 n_1 = 0 $

This will give: $n_1 = 0$ or $n_1 = - \dfrac{2}{3} $

Combining the two cases, we have three possibilities for vector $n$

1. $ n_1 = 0$ , $ n = (0, 0, -1) $

2. $ n_1 = 0$ , $ n = (0, 0, 1 ) $

3. $ n_1 = \dfrac{2}{3} $ , $ n = (\dfrac{2}{3}, - \dfrac{2}{3} , - \dfrac{1}{3} ) $

4. $ n_1 = - \dfrac{2}{3}$ , $ n = ( - \dfrac{2}{3} , \dfrac{2}{3} , \dfrac{1}{3} ) $

Note that (1) and (2) are equivalent, and that (3) and (4) are also equivalent.

Therefore, there are only two possible planes that satisfy the conditions, and they are

$ (0, 0, 1) \cdot ( r - (1, 0, 1) ) = 0 $

which is $ z = 1 $

and

$ (2, -2, -1) \cdot (r - (1, 0, 1) ) = 0 $

which is $2 (x - 1) - 2 y - (z - 1) = 0 $ i.e. $ 2 x - 2 y - z - 1 = 0 $

Answer 3

Let $\pi$ be a plane whose distance from point $C(5,0,-1)$ is $4$, then there exists a unique point $P\in\pi$ whose distance from point $C(5,0,-1)$ is $4$.

If $P(a,b,c)$, then $$\begin{cases} &\vec{AP}=(a-1,b,c-1);\\ &\vec{BP}=(a,b+1,c-1);\\ &\vec{CP}=(a-5,b,c+3). \end{cases}$$ Since $$\begin{cases} &\vec{CP}\cdot\vec{AP}=0;\\ &\vec{CP}\cdot\vec{BP}=0;\\ &|\vec{CP}|^2=16, \end{cases}$$ we have $$\begin{cases} &(a-5)(a-1)+b^2+(c+3)(c-1)=0;\\ &(a-5)a+b(b+1)+(c+3)(c-1)=0;\\ &(a-5)^2+b^2+(c+3)^2=16. \end{cases}$$ The solutions for this system of equations are $$\begin{cases} &a=5,b=0,\text{and} \ c=1;\\ &a=\frac{7}{3},b=\frac{8}{3},\text{and} \ c=-\frac{5}{3}. \end{cases}$$ Case 1: If $a=5,b=0,\text{and} \ c=1$, then $P(5,0,1)$. Therefore, it is not hard to prove that the equation of a plane passing through points $P(5,0,1),A(1,0,1),\text{and} \ B(0,-1,1)$ is given by $$\boxed{z-1=0}.$$ Case 2: If $a=\frac{7}{3},b=\frac{8}{3},\text{and} \ c=-\frac{5}{3}$, then $P(\frac{7}{3},\frac{8}{3},-\frac{5}{3})$. Therefore, it is not hard to prove that the equation of a plane passing through points $P(\frac{7}{3},\frac{8}{3},-\frac{5}{3}),A(1,0,1),\text{and} \ B(0,-1,1)$ is given by $$\boxed{2x-2y-z-1=0}.$$