Question

# 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}

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 $x-y+z=16$.

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.