Question

A line passes through (2, -1, 3) and it is perpendicular to the lines $\overrightarrow{r}=\left( \widehat{i}+\widehat{j}-\widehat{k} \right)+\lambda \left( 2\widehat{i}-2\widehat{j}+\widehat{k} \right)$and $\overrightarrow{s}=\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+\mu \left( \widehat{i}+2\widehat{j}+2\widehat{k} \right)$. Obtain its equation in vector and Cartesian form.

Answer 1

Hint: First, before proceeding for this, we must assume the line ${{L}_{1}}$as $\overrightarrow{r}=\left( \widehat{i}+\widehat{j}-\widehat{k} \right)+\lambda \left( 2\widehat{i}-2\widehat{j}+\widehat{k} \right)$ and ${{L}_{2}}$as $\overrightarrow{s}=\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+\mu \left( \widehat{i}+2\widehat{j}+2\widehat{k} \right)$and also we assume the line L which is perpendicular to line ${{L}_{1}}$,${{L}_{2}}$and its coordinates as (a, b, c). Then, if the line L is perpendicular to line${{L}_{1}}$, then we get the dot product of their vectors as zero and similar we get for line ${{L}_{2}}$. Then, by using substitution, we get the values of direction ratio of line L which gives us the vector and Cartesian equation of that line.

Complete step-by-step answer:
In this question, we are supposed to find the equation in vector and Cartesian form of a line passes through (2, -1, 3) and it is perpendicular to the lines $\overrightarrow{r}=\left( \widehat{i}+\widehat{j}-\widehat{k} \right)+\lambda \left( 2\widehat{i}-2\widehat{j}+\widehat{k} \right)$and $\overrightarrow{s}=\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+\mu \left( \widehat{i}+2\widehat{j}+2\widehat{k} \right)$.
So, before proceeding for this, we must assume the line ${{L}_{1}}$as $\overrightarrow{r}=\left( \widehat{i}+\widehat{j}-\widehat{k} \right)+\lambda \left( 2\widehat{i}-2\widehat{j}+\widehat{k} \right)$ and ${{L}_{2}}$as $\overrightarrow{s}=\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+\mu \left( \widehat{i}+2\widehat{j}+2\widehat{k} \right)$and also we assume the line L which is perpendicular to line ${{L}_{1}}$and ${{L}_{2}}$.
Then, we assume the coordinates of the line L as (a, b, c).
Now, by using the concept of vector for line L where p is any constant, we get:
$\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+p\left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right)$
Now, if the line L is perpendicular to line${{L}_{1}}$, then we get the dot product of their vectors as zero as:
$2a-2b+c=0....\left( i \right)$
Similarly, if the line L is perpendicular to line${{L}_{2}}$, then we get the dot product of their vectors as zero as:
$a+2b+2c=0....\left( ii \right)$
Now, by calculating the value of 2b from equation (i), we get:
$2b=2a+c$
Then, by substituting the value of 2b in equation (ii), we get:
$\begin{align}
  & a+2a+c+2c=0 \\
 & \Rightarrow 3a+3c=0 \\
 & \Rightarrow a=-c \\
\end{align}$
Then, by substituting the value of value of c in equation (ii), we get:
$\begin{align}
  & a+2b-2a=0 \\
 & \Rightarrow a=2b \\
 & \Rightarrow b=\dfrac{a}{2} \\
\end{align}$
So, from the above relations we get the value of coordinates of line L in terms of a as:
$\left( a,\dfrac{a}{2},-a \right)$
So, from the above if we take direction ratios of the line L, we get:
$\left( 1,\dfrac{1}{2},-1 \right)$
So, we get the equation of L in vector form is written as:
$\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+p\left( \widehat{i}+\dfrac{\widehat{j}}{2}-\widehat{k} \right)$
Now, we know that the Cartesian form of the equation can be written where $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ are the values from the first vector of line L as:
$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$
So, by substituting all the values from the vector equation, we get the Cartesian equation as:
$\dfrac{x-2}{1}=\dfrac{y+1}{\dfrac{1}{2}}=\dfrac{z-3}{-1}$
Hence, the vector equation and Cartesian equation of the line is $\left( 2\widehat{i}-\widehat{j}-3\widehat{k} \right)+p\left( \widehat{i}+\dfrac{\widehat{j}}{2}-\widehat{k} \right)$ and $\dfrac{x-2}{1}=\dfrac{y+1}{\dfrac{1}{2}}=\dfrac{z-3}{-1}$ respectively.

Note: Now, to solve this type of questions we must know the angle condition of the direction ratio’s which is used above in the question as both the lines are perpendicular so the dot product is zero is given by the condition that:
$\cos \theta =\dfrac{\left| {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}} \right|}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}+{{c}_{1}}^{2}}\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}+{{c}_{2}}^{2}}}$
Where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are the points of the line.