Question
Answers

The distance of the point \[\left( 1,0,2 \right)\] from the point of intersection of the line \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] and the plane \[x-y+z=16\], is
\[\begin{align}
  & A)2\sqrt{14} \\
 & B)8 \\
 & C)3\sqrt{21} \\
 & D)13 \\
\end{align}\]

Answer Verified Verified
Hint: Let us assume \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] is equal to k. Now we should find the values of x, y and z in terms of k. Now we should put these values in the plane \[x-y+z=16\]. Now we will get the value of k. From this, we can find the values of x, y and z. We know that the distance between \[A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\] is equal to\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\]. So, in this way we can find the distance of the point \[\left( 1,0,2 \right)\] from the point of intersection of the line \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] and the plane \[x-y+z=16\].

Complete step-by-step answer:
Let us assume \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] is equal to k.
Then let us consider
\[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}=k........(1)\]
From equation (1), we can write
\[\Rightarrow \dfrac{x-2}{3}=k\]
Now by using cross multiplication, then we get
\[\begin{align}
  & \Rightarrow x-2=3k \\
 & \Rightarrow x=3k+2....(2) \\
\end{align}\]
From equation (1), we can write
\[\Rightarrow \dfrac{y+1}{4}=k\]
Now by using cross multiplication, then we get
\[\begin{align}
  & \Rightarrow y+1=4k \\
 & \Rightarrow y=4k-1....(3) \\
\end{align}\]
From equation (1), we can write
\[\Rightarrow \dfrac{z-1}{12}=k\]
Now by using cross multiplication, then we get
\[\begin{align}
  & \Rightarrow z-1=12k \\
 & \Rightarrow x=12k+1....(4) \\
\end{align}\]
From the question, it is given that the equation of the plane is \[x-y+z=16\].
Let us consider
\[x-y+z=16....(5)\]
Now we should find the point of intersection of the line \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] and the plane \[x-y+z=16\].
Now, let us substitute equation (2), equation (3) and equation (4) in equation (5), then we get
\[\begin{align}
  & \Rightarrow 3k+2-4k+1+12k+2=16 \\
 & \Rightarrow 11k+5=11 \\
 & \Rightarrow k=1....(6) \\
\end{align}\]
Now let us substitute equation (6) in equation (2), then we get
\[\begin{align}
  & \Rightarrow x=3(1)+2 \\
 & \Rightarrow x=3+2 \\
 & \Rightarrow x=5.....(7) \\
\end{align}\]
Now let us substitute equation (6) in equation (3), then we get
\[\begin{align}
  & \Rightarrow y=4(1)-1 \\
 & \Rightarrow y=4-1 \\
 & \Rightarrow y=3.....(8) \\
\end{align}\]
Now let us substitute equation (6) in equation (4), then we get
\[\begin{align}
  & \Rightarrow z=12\left( 1 \right)+2 \\
 & \Rightarrow z=12+2 \\
 & \Rightarrow z=14.....(9) \\
\end{align}\]
So, the point of intersection of the line \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] and the plane \[x-y+z=16\], is \[\left( 5,3,14 \right)\].
Now we should find the distance between \[\left( 5,3,14 \right)\] and \[\left( 1,0,2 \right)\].
We know that the distance between \[A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\] is equal to\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\].
Let us assume the distance between \[\left( 5,3,14 \right)\] and \[\left( 1,0,2 \right)\] is equal to d.
\[\begin{align}
  & \Rightarrow d=\sqrt{{{\left( 1-5 \right)}^{2}}+{{\left( 0-3 \right)}^{2}}+{{\left( 2-14 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( -12 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{16+9+144} \\
 & \Rightarrow d=\sqrt{169} \\
 & \Rightarrow d=13.....(10) \\
\end{align}\]
So, from equation (10) it is clear that the distance of the point \[\left( 1,0,2 \right)\] from the point of intersection of the line \[\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}\] and the plane \[x-y+z=16\], is equal to 13.

So, the correct answer is “Option D”.

Note: Students may have a misconception that the distance between \[A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and \[B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\] is equal to\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}-{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}\]. If this misconception is followed, then we cannot get the correct measure of distance. So, students should avoid these mistakes and misconceptions.