Question

Find the vector and Cartesian equation of the line passing through the point (2, 1, 3) and perpendicular to the lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ and $\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}$.

Answer 1

We know that the equation of any line in vector form is represented as $\overrightarrow{a}+\lambda \overrightarrow{b}$, where $\vec{a}$ is the vector of the point through which line is passing, $\lambda$ is a constant and $\vec{b}$is a unit vector parallel to the line. So, we need to find the vector $\vec{b}$. It is given that the vector is perpendicular to lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ and $\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}$. Based on this information, we will find the direction ratios of vectors $\vec{b}$. Once we find $\vec{b}$, we will be able to find the equation of line in vector form. Equation of line in standard is given as $\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$, where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is the point through which the line is passing and a, b, c are the direction ratios of the line.

Complete step by step answer:
Let $\vec{r}=\vec{a}+\lambda \overrightarrow{b}$ be the equation of the line in standard form, where $\vec{a}$ is the vector of the point through which line is passing, $\lambda$ is a constant and $\vec{b}$is a unit vector parallel to the line.
Since the line is passing through (2, 1, 3), $\vec{a}=2\hat{i}+\hat{k}+3\hat{j}$
Now, it is given that the line is perpendicular to $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ and $\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}$.
We will first convert these lines in the vector form.
The first line in vector form will be $\hat{i}+2\hat{j}+3\hat{k}+\lambda \left( \hat{i}+2\hat{j}+3\hat{k} \right)$ and second line will be $\left( 0 \right)\hat{i}+\left( 0 \right)\hat{j}+\left( 0 \right)\hat{k}+\lambda \left( -3\hat{i}+2\hat{j}+5\hat{k} \right)$
Thus, the vector $\vec{b}$ will be perpendicular to $\hat{i}+2\hat{j}+3\hat{k}$ and $-3\hat{i}+2\hat{j}+5\hat{k}$. To find the vector $\vec{b}$, we will find the cross product of $\hat{i}+2\hat{j}+3\hat{k}$ and $-3\hat{i}+2\hat{j}+5\hat{k}$
If two vectors are ${{r}_{1}}=a\hat{i}+b\hat{j}+c\hat{k}$ and ${{r}_{2}}=p\hat{i}+q\hat{j}+r\hat{k}$, then the cross product of the two vectors will be represented by ${{r}_{1}}\times {{r}_{2}}$ and we can find it by finding the determinant $\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ a & b & c \\\ p & q & r \\\ \end{matrix} \right|$
Therefore, we will find the cross product as $\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 2 & 3 \\\ -3 & 2 & 5 \\\ \end{matrix} \right|$.
Thus, the vector $\vec{b}$is $4\hat{i}-14\hat{j}+8\hat{k}$.
The equation of the line is as follows:
$\begin{aligned} & \Rightarrow \vec{r}=2\hat{i}+\hat{k}+3\hat{j}+\lambda \left( 4\hat{i}-14\hat{j}+8\hat{k} \right) \\\ & \Rightarrow \vec{r}=2\hat{i}+\hat{k}+3\hat{j}+2\lambda \left( 2\hat{i}-7\hat{j}+4\hat{k} \right) \\\ & \Rightarrow \vec{r}=2\hat{i}+\hat{k}+3\hat{j}+t\left( 2\hat{i}-7\hat{j}+4\hat{k} \right) \\\ \end{aligned}$
Where t is a constant. The directional ratios of this line will be (2, ─7, 4).

Therefore, the equation of the line in cartesian form is $\dfrac{x-2}{2}=\dfrac{y-1}{-7}=\dfrac{z-3}{4}$.

Note: Students can note that the sum of the products of the DRs of perpendicular lines is 0. Thus, if a, b, c are the DRs of the required lines which is perpendicular to lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ and $\dfrac{x}{-3}=\dfrac{y}{2}=\dfrac{z}{5}$, whose DRs will be (1, 2, 3) and (─3, 2, 5) respectively, then (a + 2b + 3c = 0) and (─3a + 2b + 5c = 0). By solving these two equations, we can get $\dfrac{a}{4}=\dfrac{b}{-14}=\dfrac{c}{8}$. Thus, we find the DRs in this way.