Question

The equation of the straight line passing through the origin and perpendicular to the lines $\dfrac{x+1}{-3}=\dfrac{y-2}{2}=\dfrac{z}{1}$ and $\dfrac{x-1}{1}=\dfrac{y}{-3}=\dfrac{z+1}{2}$has the equation. \[\]
A. $x=y=z$\[\]
B. $\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{6}$\[\]
C. $\dfrac{x}{3}=\dfrac{y}{1}=\dfrac{z}{0}$\[\]
D. None of these \[\]

Answer 1

Hint: We convert the given equation of line in component vector form$\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}}$. We find the vector perpendicular to $\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}}$ by taking the cross product of them that is$\overrightarrow{r}=\overrightarrow{{{r}_{1}}}\times \overrightarrow{{{r}_{2}}}$. We find the direction ratio from the vector $\left( a,b,c \right)$ and then use equation line in a space passing through a point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$.\[\]

Complete step-by-step answer:
We know that the direction ratios are any three numbers which are proportional to direction cosines. The equation of any line passing through a point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and direction ratios $\left( a,b,c \right)$ is given by
\[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
We can convert any line in above form to the component vector form say $\overrightarrow{r}$ as
\[\overrightarrow{r}=a\hat{i}+b\hat{j}+c\hat{k}\]
Here $\hat{i},\hat{j},\hat{k}$ are orthogonal unit vectors and $ka,kb,kc$ are called components for any integer$k$. We know that if two vectors $\overrightarrow{{{r}_{1}}}={{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k}$ and $\overrightarrow{{{r}_{2}}}={{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k}$ lie on one plane then the vector perpendicular to both of them is given by the cross product of $\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}}$ that is
\[\overrightarrow{{{r}_{1}}}\times \overrightarrow{{{r}_{2}}}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|\]

We are given in the question two equations of lines as
\[\begin{align}
  & \dfrac{x+1}{-3}=\dfrac{y-2}{2}=\dfrac{z}{1}....\left( 1 \right) \\
 & \dfrac{x-1}{1}=\dfrac{y}{-3}=\dfrac{z+1}{2}.....\left( 2 \right) \\
\end{align}\]
Let us convert line (1) and (2) to the corresponding component vector forms denoted by $\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}}$ as
\[\begin{align}
  & \overrightarrow{{{r}_{1}}}=-3\hat{i}+2\hat{j}+\hat{k} \\
 & \overrightarrow{{{r}_{2}}}=\hat{i}-3\hat{j}+2\hat{k} \\
\end{align}\]
We are going to find the vector perpendicular to the vectors $\overrightarrow{{{r}_{1}}},\overrightarrow{{{r}_{2}}}$ denoted by $\overrightarrow{r}$ using the determinant for cross product as
\[\begin{align}
  & \overrightarrow{r}=\overrightarrow{{{r}_{1}}}\times \overrightarrow{{{r}_{2}}} \\
 & \Rightarrow \overrightarrow{r}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   -3 & 2 & 1 \\
   1 & -3 & 2 \\
\end{matrix} \right| \\
\end{align}\]
We expand by the first row and have,
\[\begin{align}
  & \Rightarrow \overrightarrow{r}=\hat{i}\left( 4-\left( -3 \right) \right)-\hat{j}\left( -6-1 \right)+\hat{k}\left( 9-2 \right) \\
 & \Rightarrow \overrightarrow{r}=7\hat{i}+7\hat{j}+7\hat{k} \\
\end{align}\]
We divide the components by 7 and find the direction of the line perpendicular to both the line (1) and (2) as $\left( 1,1,1 \right)$ since all direction ratios of a line are proportional to each other. We are also given that the perpendicular line passes through the origin. So we have the equation of perpendicular line origin $\left( 0,0,0 \right)$as the point and direction ratios $\left( 1,1,1 \right)$ as
\[\begin{align}
  & \dfrac{x-0}{1}=\dfrac{y-0}{1}=\dfrac{z-0}{1} \\
 & \Rightarrow x=y=z \\
\end{align}\]
So the correct option is A. The rough figure of the problem is drawn below. \[\]

So, the correct answer is “Option A”.

Note: We note that the direction of the vector $\overrightarrow{r}$ is determined by right hand thumb rule and line does not have any direction. All the lines parallel to a line with direction ratios $\left( a,b,c \right)$ will have direction ratio $\left( ka,kb,kc \right)$ for some non-zero integer$k$. We can alternatively solve using the relation between direction ratios of two perpendicular lines that is ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$.