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

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

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