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

& A)2\sqrt{14} \\\ & B)8 \\\ & C)3\sqrt{21} \\\ & D)13 \\\ \end{aligned}$$

Answer 1

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

& \Rightarrow x-2=3k \\\ & \Rightarrow x=3k+2....(2) \\\ \end{aligned}$$ From equation (1), we can write $$\Rightarrow \dfrac{y+1}{4}=k$$ Now by using cross multiplication, then we get $$\begin{aligned} & \Rightarrow y+1=4k \\\ & \Rightarrow y=4k-1....(3) \\\ \end{aligned}$$ From equation (1), we can write $$\Rightarrow \dfrac{z-1}{12}=k$$ Now by using cross multiplication, then we get $$\begin{aligned} & \Rightarrow z-1=12k \\\ & \Rightarrow x=12k+1....(4) \\\ \end{aligned}$$ From the question, it is given that the equation of the plane is $$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{aligned} & \Rightarrow 3k+2-4k+1+12k+2=16 \\\ & \Rightarrow 11k+5=11 \\\ & \Rightarrow k=1....(6) \\\ \end{aligned}$$ Now let us substitute equation (6) in equation (2), then we get $$\begin{aligned} & \Rightarrow x=3(1)+2 \\\ & \Rightarrow x=3+2 \\\ & \Rightarrow x=5.....(7) \\\ \end{aligned}$$ Now let us substitute equation (6) in equation (3), then we get $$\begin{aligned} & \Rightarrow y=4(1)-1 \\\ & \Rightarrow y=4-1 \\\ & \Rightarrow y=3.....(8) \\\ \end{aligned}$$ Now let us substitute equation (6) in equation (4), then we get $$\begin{aligned} & \Rightarrow z=12\left( 1 \right)+2 \\\ & \Rightarrow z=12+2 \\\ & \Rightarrow z=14.....(9) \\\ \end{aligned}$$ So, the point of intersection of the line $$\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{aligned} & \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{aligned}$$ So, from equation (10) it is clear that the distance of the point $$\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.