Question

Find the equation of the plane determined by the points $A\left( 3,-1,2 \right),B\left( 5,2,4 \right),C\left( -1,-1,6 \right)$ and hence find the distance between the plane and the point$P\left( 6,5,9 \right)$.

Answer 1

Hint: To solve this question, we should find the directional ratios of the normal vector to the plane. The lines AB and BC are perpendicular to the normal of the place. We know that the directional ratios of points $D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),E\left( {{x}_{^{2}}},{{y}_{2}},{{z}_{2}} \right)$ are given by ${{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}}$ respectively. Using this we can find the directional ratios of the lines AB and BC. We know that the condition for two lines with directional ratios $a,b,c$ and ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ is given by $a{{a}_{1}}+b{{b}_{1}}+c{{c}_{1}}=0$. By assuming the directional ratios of the normal are $a,b,c$ and using the above perpendicular relation with AB and BC, we get the directional ratios of the normal of the plane. The equation of the plane with directional ratios of the normal as $a,b,c$ is given by $ax+by+cz=d$ where d is a constant. We can write the equation of the plane by assuming a constant d. By substituting either of the points A, B, C we get the value of d and the equation of the plane. The perpendicular distance from a point $D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ to the plane $ax+by+cz=d$is given by $\text{distance}=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}-d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. By using this formula, we can get the distance between the point P and the plane.

Complete step by step answer:
We know that the directional ratios of points $D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),E\left( {{x}_{^{2}}},{{y}_{2}},{{z}_{2}} \right)$ are given by ${{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}}$ respectively.
We can calculate the directional ratios of the lines AB and BC using the above formula.
Directional ratios of AB where $A\left( 3,-1,2 \right),B\left( 5,2,4 \right)$ are
$\begin{align}
  & d.r's=3-5,-1-2,2-4 \\
 & d.r's=-2,-3,-2 \\
\end{align}$
Directional ratios of BC where $B\left( 5,2,4 \right),C\left( -1,-1,6 \right)$ are
$\begin{align}
  & d.r's=5-\left( -1 \right),2-\left( -1 \right),4-6 \\
 & d.r's=6,3,-2 \\
\end{align}$
Let us consider that the directional ratios of the normal of the plane be $a,b,c$
We know that the condition for two lines with directional ratios $a,b,c$ and ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ is given by $a{{a}_{1}}+b{{b}_{1}}+c{{c}_{1}}=0$.
We know that the normal line is perpendicular to both the lines AB and BC.
Using this condition on normal and the line AB, we get
$\begin{align}
  & -2a-3b-2c=0 \\
 & 2a+3b+2c=0\to \left( 1 \right) \\
\end{align}$
Using this condition on normal and the line BC, we get
$6a+3b-2c=0\to \left( 2 \right)$
Let us substitute $c=1$ in both the equations-1 and 2, we get
$\begin{align}
  & 2a+3b=-2\to \left( 3 \right) \\
 & 6a+3b=2\to \left( 4 \right) \\
\end{align}$
By solving the two equations, we get the values of a, b.
$\left( 4 \right)-\left( 3 \right)$, we get
$\begin{align}
  & 4a=4 \\
 & a=1 \\
\end{align}$
Using a = 1 in equation-3, we get
$\begin{align}
  & 2+3b=-2 \\
 & 3b=-4 \\
 & b=\dfrac{-4}{3} \\
\end{align}$
The directional ratios of the normal are
$1,\dfrac{-4}{3},1$
As they are ratios, we can multiply them by 3
The final directional ratios are $3,-4,3$
The equation of the plane with directional ratios of the normal as $a,b,c$ is given by $ax+by+cz=d$
We can get the required plane equation as
$3x-4y+3z=d$ where d is a constant
By substituting the point $B\left( 5,2,4 \right)$, we get
$\begin{align}
  & 3\times 5-4\times 2+3\times 4=d \\
 & d=27-8=19 \\
\end{align}$
The equation of the plane is $3x-4y+3z=19$
The distance from a point $D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ to the plane $ax+by+cz=d$is given by $\text{distance}=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}-d \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$.
We can get the distance between the point $P\left( 6,5,9 \right)$ from $3x-4y+3z=19$ as
$\text{distance}=\dfrac{\left| 3\times 6-4\times 5+3\times 9-19 \right|}{\sqrt{{{3}^{2}}+{{4}^{2}}+{{3}^{2}}}}=\dfrac{\left| 18+27-20-19 \right|}{\sqrt{34}}=\dfrac{6}{\sqrt{34}}$
$\therefore $The equation of the plane is $3x-4y+3z=19$ and the distance of the point $P\left( 6,5,9 \right)$ from the plane is $\dfrac{6}{\sqrt{34}}$

Note:
An alternate way to do this problem by considering the equation of the plane as
$ax+by+cz=1$ and substituting the three points A, B, C to get three different equations in a, b, c. We can write them as
$\begin{align}
  & 3a-b+2c=1 \\
 & 5a+2b+4c=1 \\
 & -a-b+6c=1 \\
\end{align}$
By solving these equations, we can get the values of a, b, c and the required equation of the plane.