Question

The shortest distance between the lines $\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}$ is?
(a) $\sqrt{30}$
(b) $2\sqrt{30}$
(c) $5\sqrt{30}$
(d) $3\sqrt{30}$

Answer 1

Hint: Consider the two lines of the form $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$. Now, compare the values of $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ which are considered as the direction ratios of the two lines. Now, write then in vector form as ${{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}$ and ${{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}$. Assume the shortest distance between the lines in vector form as $x\hat{i}+y\hat{j}+z\hat{k}$ and take the dot product of this vector with the two vectors assumed above one by one and equate them with 0 to form two linear equations in three variables x, y and z. Use the cross multiplication method for solving linear equations and write the relation in the form $\dfrac{x}{m}=\dfrac{y}{n}=\dfrac{z}{p}$. Use the formula d = $\sqrt{{{m}^{2}}+{{n}^{2}}+{{p}^{2}}}$ to calculate the shortest distance. Here, d represents the shortest distance.

Complete step by step solution:
Here we have been provided with two lines $\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}$ and we are asked to find the shortest distance between the two.
Now, assuming the two lines of the form $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ we know that the direction ratios of these lines is given as $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ respectively. So we have $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)=\left( 3,-1,1 \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)=\left( -3,2,4 \right)$. In vector form we can write them as $3\hat{i}-\hat{j}+\hat{k}$ and $-3\hat{i}+2\hat{j}+4\hat{k}$ respectively.
Let us assume that the shortest distance between the two lines is given as $x\hat{i}+y\hat{j}+z\hat{k}$ in vector form. Now, we know that the shortest distance between the two lines is the perpendicular distance between them. That means the vector $x\hat{i}+y\hat{j}+z\hat{k}$ is perpendicular to the vectors $3\hat{i}-\hat{j}+\hat{k}$ and $-3\hat{i}+2\hat{j}+4\hat{k}$. This leads us to the conclusion that their dot product will be equal to 0. So considering the dot products one by one we get,
(1) $\left( x\hat{i}+y\hat{j}+z\hat{k} \right).\left( 3\hat{i}-\hat{j}+\hat{k} \right)=0$ so we get,
$\Rightarrow 3x-y+z=0$……… (i)
(2) $\left( x\hat{i}+y\hat{j}+z\hat{k} \right).\left( -3\hat{i}+2\hat{j}+4\hat{k} \right)=0$ so we get,
$\Rightarrow -3x+2y+4z=0$ ……… (ii)
We know that the linear equations of the form ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z=0$ can be simplified using the cross multiplication method as $\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{z}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$ so substituting the values we get,
$\begin{align}
  & \Rightarrow \dfrac{x}{-4-2}=\dfrac{y}{-3-12}=\dfrac{z}{6-3} \\
 & \Rightarrow \dfrac{x}{-6}=\dfrac{y}{-15}=\dfrac{z}{3} \\
\end{align}$
So the direction ratio of the line of the shortest distance is (-6, -15, 3). Therefore the shortest distance (d) will be given as: -
$\begin{align}
  & \Rightarrow d=\sqrt{{{\left( -6 \right)}^{2}}+{{\left( -15 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{36+225+9} \\
 & \Rightarrow d=\sqrt{270} \\
 & \therefore d=3\sqrt{30} \\
\end{align}$

So, the correct answer is “Option d”.

Note: You can remember a direct formula to find the shortest distance between the two lines $\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ given as $d=\left| \dfrac{\left( \begin{matrix}
   {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right)}{\sqrt{{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}~}} \right|$. This formula will help in solving the question in less time but you have to be careful in calculations as it is a length formula with many terms so the possibility of making small calculation mistakes is high.