Question

Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+3\widehat{k} \right)-4=0$ and $\overrightarrow{r}.(-2\widehat{i}+\widehat{j}+\widehat{k})+5=0$ and whose intercept on $X-axis$ is equal to that of on $Y-axis$.

Answer 1

Hint: Here in this question we need to find the equation of the plane consisting the line of intersection of both the planes $\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+3\widehat{k} \right)-4=0$ and $\overrightarrow{r}.(-2\widehat{i}+\widehat{j}+\widehat{k})+5=0$. From the concept we know that when $\overrightarrow{r}.\widehat{{{n}_{1}}}={{d}_{1}}$ is the equation of a plane, then the position vector of any point $\overrightarrow{t}$ on the line intersecting the given plane must satisfy the equation, that is $\overrightarrow{t}.\widehat{{{n}_{1}}}={{d}_{1}}$.

Complete step by step answer:
We have equations of 2 planes $\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+3\widehat{k} \right)-4=0$ and $\overrightarrow{r}.(-2\widehat{i}+\widehat{j}+\widehat{k})+5=0$.
We need to find the equation of the plane consisting of the line of intersection of both the planes.
We know that it will be $X+\lambda Y=0$ where $X,Y$ are the two planes and $\lambda $ is a constant.
$\overrightarrow{r}.\left( \widehat{i}\left( 1-2\lambda \right)+\widehat{j}\left( \lambda -2 \right)+\widehat{k}\left( \lambda +3 \right) \right)-4+5\lambda =0$
We can consider $\left( 1-2\lambda \right)x+\left( \lambda -2 \right)y+\left( 3+\lambda \right)z=4-5\lambda $
 By simplifying this,
$\dfrac{x}{\left( \dfrac{4-5\lambda }{1-2\lambda } \right)}+\dfrac{y}{\left( \dfrac{4-5\lambda }{\lambda -2} \right)}+\dfrac{z}{\left( \dfrac{4-5\lambda }{3+\lambda } \right)}=1$
From the question, we know that the intercept on $X-axis$ is equal to that of on $Y-axis$.
So,\[\begin{align}
  & \left( \dfrac{4-5\lambda }{1-2\lambda } \right)=\left( \dfrac{4-5\lambda }{\lambda -2} \right) \\
 & 1-2\lambda =\lambda -2 \\
 & 3=3\lambda \\
 & \lambda =1 \\
\end{align}\]
Here the value of $\lambda $ is 1.
So now by substituting the value of $\lambda $ we have $x+y-4z-1=0$
   From this the equation of the plane consisting of the line of intersection of both the planes will be $\overrightarrow{r}.\left( \widehat{i}+\widehat{j}-4\widehat{k} \right)-1=0$.

Note: From the concept we know that when $\overrightarrow{r}.\widehat{{{n}_{1}}}={{d}_{1}}$and$\overrightarrow{r}.\widehat{{{n}_{2}}}={{d}_{2}}$ are the equations of the given two planes respectively, then the position vector of any point $\overrightarrow{t}$ on the line of intersection of both the planes must satisfy the given plane equations, that is $\overrightarrow{t}.\widehat{{{n}_{1}}}={{d}_{1}}$and$\overrightarrow{t}.\widehat{{{n}_{2}}}={{d}_{2}}$. By combining these two equations we get $\overrightarrow{t}.\left( \widehat{{{n}_{1}}}+\lambda \widehat{{{n}_{2}}} \right)={{d}_{1}}+\lambda {{d}_{2}}$.
Hence, from the equation $\overrightarrow{t}.\left( \widehat{{{n}_{1}}}+\lambda \widehat{{{n}_{2}}} \right)={{d}_{1}}+\lambda {{d}_{2}}$ we get the equation of the plane consisting the line of intersection of both the planes $\overrightarrow{r}.\left( \widehat{{{n}_{1}}}+\lambda \widehat{{{n}_{2}}} \right)={{d}_{1}}+\lambda {{d}_{2}}$.