Question

Find the equation of a line passing through the point $P\left( 2,-1,3 \right)$ and perpendicular to the lines \[=\hat{i}+\hat{j}-\hat{k}+\lambda \left( 2\hat{i}-2\hat{j}+\hat{k} \right)\] and $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}-3\hat{k} \right)+\mu \left( \hat{i}+2\hat{j}+2\hat{k} \right)$?

Answer 1

Hint: We start solving the problem by recalling the equation of the line passing through point $\left( a,b,c \right)$ and parallel to the vector $d\hat{i}+e\hat{j}+f\hat{k}$ is $\overrightarrow{r}=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)+\alpha \left( d\hat{i}+e\hat{j}+f\hat{k} \right)$. We assume the vector parallel to the required line and write the equation of it. We can then see that the given lines are parallel to two vectors. We then find the perpendicular vector to both the vectors and replace it in the equation of the line to get the required result.

Complete step-by-step solution:
According to the problem, we need to find the equation of a line which is passing through the point $P\left( 2,-1,3 \right)$ and perpendicular to the lines \[=\hat{i}+\hat{j}-\hat{k}+\lambda \left( 2\hat{i}-2\hat{j}+\hat{k} \right)\] and $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}-3\hat{k} \right)+\mu \left( \hat{i}+2\hat{j}+2\hat{k} \right)$.
Let us assume \[=\hat{i}+\hat{j}-\hat{k}+\lambda \left( 2\hat{i}-2\hat{j}+\hat{k} \right)\] as ${{L}_{1}}$ and $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}-3\hat{k} \right)+\mu \left( \hat{i}+2\hat{j}+2\hat{k} \right)$ as ${{L}_{2}}$. Let us also assume the required line be ${{L}_{3}}$ and the vector parallel to the line be $p\hat{i}+q\hat{j}+r\hat{k}$.
We know that the vector equation of the line passing through point $\left( a,b,c \right)$ and parallel to the vector $d\hat{i}+e\hat{j}+f\hat{k}$ is $\overrightarrow{r}=\left( a\hat{i}+b\hat{j}+c\hat{k} \right)+\alpha \left( d\hat{i}+e\hat{j}+f\hat{k} \right)$, where $\alpha $is an arbitrary constant. Using this fact, we can see that ${{L}_{1}}$ is parallel to the vector \[2\hat{i}-2\hat{j}+\hat{k}\] and ${{L}_{2}}$ is parallel to the vector $\hat{i}+2\hat{j}+2\hat{k}$.
According the problem, the line ${{L}_{3}}$ passes through the point $P\left( 2,-1,3 \right)$, so we get the equation of the line as $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}+3\hat{k} \right)+\beta \left( p\hat{i}+q\hat{j}+r\hat{k} \right)$.
perpendicular to both ${{L}_{1}}$ and ${{L}_{2}}$. So, this makes that the line is parallel to the vector that is perpendicular to the vectors \[2\hat{i}-2\hat{j}+\hat{k}\] and $\hat{i}+2\hat{j}+2\hat{k}$.
We know that the vector perpendicular to the vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ is $\overrightarrow{x}\times \overrightarrow{y}$. We know that the cross product of two vectors $a\hat{i}+b\hat{j}+c\hat{k}$ and $d\hat{i}+e\hat{j}+f\hat{k}$ is defined as $\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   a & b & c \\
   d & e & f \\
\end{matrix} \right|$.
So, the vector perpendicular to the vectors \[2\hat{i}-2\hat{j}+\hat{k}\] and $\hat{i}+2\hat{j}+2\hat{k}$ is $\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|=\hat{i}\times \left| \begin{matrix}
   -2 & 1 \\
   2 & 2 \\
\end{matrix} \right|-\hat{j}\times \left| \begin{matrix}
   2 & 1 \\
   1 & 2 \\
\end{matrix} \right|+\hat{k}\left| \begin{matrix}
   2 & -2 \\
   1 & 2 \\
\end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|=\hat{i}\times \left( \left( -2\times 2 \right)-\left( 2\times 1 \right) \right)-\hat{j}\times \left( \left( 2\times 2 \right)-\left( 1\times 1 \right) \right)+\hat{k}\left( \left( 2\times 2 \right)-\left( 1\times -2 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|=\hat{i}\times \left( -4-2 \right)-\hat{j}\times \left( 4-1 \right)+\hat{k}\left( 4-\left( -2 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|=\hat{i}\times \left( -6 \right)-\hat{j}\times \left( 3 \right)+\hat{k}\left( 6 \right)$.
$\Rightarrow \left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 2 & 2 \\
\end{matrix} \right|=-6\hat{i}-3\hat{j}+6\hat{k}$.
The vector perpendicular to \[2\hat{i}-2\hat{j}+\hat{k}\] and $\hat{i}+2\hat{j}+2\hat{k}$ is $-6\hat{i}-3\hat{j}+6\hat{k}$. So, the line ${{L}_{3}}$ is parallel to $-6\hat{i}-3\hat{j}+6\hat{k}$.
This makes our required equation of the line as $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}+3\hat{k} \right)+\beta \left( -6\hat{i}-3\hat{j}+6\hat{k} \right)$.
∴ The equation of the line passing through the point $P\left( 2,-1,3 \right)$ and perpendicular to the lines \[=\hat{i}+\hat{j}-\hat{k}+\lambda \left( 2\hat{i}-2\hat{j}+\hat{k} \right)\] and $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}-3\hat{k} \right)+\mu \left( \hat{i}+2\hat{j}+2\hat{k} \right)$ is $\overrightarrow{r}=\left( 2\hat{i}-\hat{j}+3\hat{k} \right)+\beta \left( -6\hat{i}-3\hat{j}+6\hat{k} \right)$.

Note: We can also find the vector perpendicular to the vectors \[2\hat{i}-2\hat{j}+\hat{k}\] and $\hat{i}+2\hat{j}+2\hat{k}$ by using the fact that the dot product of two perpendicular vectors is zero. We can also use cartesian form to solve this form which will be easier to imagine. Whenever we are unable to do proceed in the vector form, we can convert it to cartesian form (3-d equations) and proceed further to get a clear view of the equations.