Question

The points \[A\left( -1,3,0 \right)\], \[B\left( 2,2,1 \right)\]and \[C\left( 1,1,3 \right)\] determine a plane. The distance of the plane \[ABC\] from the point \[D\left( 5,7,8 \right)\] is:
(a) \[\sqrt{66}\]
(b) \[\sqrt{71}\]
(c) \[\sqrt{73}\]
(d) \[\sqrt{76}\]

Answer 1

Hint: We will first use the formula $\left| \begin{matrix}
   x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
   {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
   {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\
\end{matrix} \right|=0$ to find the equation of the plane $ABC $ passing through these points A, B, and C since we know from the question that the points A, B, and C determine a plane. Then we will find the distance of the point $D$ from this plane $ABC$.

Complete step-by-step solution:
We know that the equation of plane passing through any three points \[\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right),\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)\] is given by
$\left| \begin{matrix}
   x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
   {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
   {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\
\end{matrix} \right|=0$
Here, from the question, we get
\[\begin{align}
  & {{x}_{1}}=-1,{{x}_{2}}=2,{{x}_{3}}=1, \\
 & {{y}_{1}}=3,{{y}_{2}}=2,{{y}_{3}}=1, \\
 & {{z}_{1}}=0,{{z}_{2}}=1,{{z}_{3}}=3 \\
\end{align}\]
Then the equation of plane $ABC$ passing through points A, B, and C is given by:
\[\left| \begin{matrix}
   x-\left( -1 \right) & y-3 & z-0 \\
   2-\left( -1 \right) & 2-3 & 1-0 \\
   1-\left( -1 \right) & 1-3 & 3-0 \\
\end{matrix} \right|=0\]
Solving it further, we get
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   x+1 & y-3 & z-0 \\
   2+1 & 2-3 & 1-0 \\
   1+1 & 1-3 & 3-0 \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   x+1 & y-3 & z \\
   3 & -1 & 1 \\
   2 & -2 & 3 \\
\end{matrix} \right|=0 \\
\end{align}\]
Expanding along the first row and solving further, we get
\[\begin{align}
  & \Rightarrow \left( x+1 \right)\left\{ \left( -1 \right)\cdot 3-1\cdot \left( -2 \right) \right\}-\left( y-3 \right)\left\{ 3\cdot 3-2\cdot 1 \right\}+z\left\{ 3\cdot \left( -2 \right)-2\left( -1 \right) \right\}=0 \\
 & \Rightarrow \left( x+1 \right)\left( -3+2 \right)-\left( y-3 \right)\left( 7 \right)+z\left( -6+2 \right)=0 \\
 & \Rightarrow -\left( x+1 \right)-7\left( y-3 \right)-4z=0 \\
 & \Rightarrow -x-1-7y+21-4z=0 \\
 & \Rightarrow -x-7y-4z+20=0 \\
 & \Rightarrow x+7y+4z-20=0 \\
\end{align}\]
We know that the distance of a point $\left( {{x}_{4}},{{y}_{4}},{{z}_{4}} \right)$ from a plane $ax+by+cz+d=0$ is given by
$T=\left| \dfrac{a{{x}_{4}}+b{{y}_{4}}+c{{z}_{4}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \right|$
Then the distance of the point D from the plane ABC is given by
$\begin{align}
& T=\left| \dfrac{5\cdot 1+7\cdot 7+8\cdot 4-20}{\sqrt{{{1}^{2}}+{{7}^{2}}+{{4}^{2}}}} \right|\\
 & =\left| \dfrac{5+49+32-20}{\sqrt{1+49+16}} \right| \\
 & =\left| \dfrac{66}{\sqrt{66}} \right| \\
 & =\sqrt{66}
\end{align}$
Hence, the correct option is (a).

Note: We should always be careful with the plus (+) and minus (-) sign while writing the formulas and calculating, because a single error can lead to change in the answer drastically. Also, practice by solving a few similar types of questions so that there is less confusion regarding the positions of $x,y,z,{{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{y}_{1}},{{y}_{2}},{{y}_{3}},{{y}_{4}},{{z}_{1}},{{z}_{2}},{{z}_{3}},{{z}_{4}}$ in the above used formulas.