Question

Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$. Also find the distance of the plane obtained above, from the origin.

Answer 1

Hint: First, before proceeding for this, we should suppose the names of the two planes as ${{P}_{1}}$ and ${{P}_{2}}$ which represents the planes $x+y+z=1$ and $2x+3y+4z=5$ respectively. Then, by using the condition for the two planes intersection where $\lambda$ is any constant as ${{P}_{2}}+\lambda {{P}_{1}}=0$. Then, we know that for the above stated plane be perpendicular to the plane $x-y+z=0$, the dot product of the normal vectors of the planes be zero and we get the value of $\lambda $. Then, by using it, we get the required plane equation. Then by using the formula as $d=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C{{z}_{1}}-D}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}} \right|$, we get the distance of the plane from origin.

Complete step-by-step answer:
In this question, we are supposed to find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$ and also find the distance of the plane obtained above, from the origin.
So, before proceeding for this, we should suppose the names of the two planes as ${{P}_{1}}$ and ${{P}_{2}}$ which represents the planes $x+y+z=1$ and $2x+3y+4z=5$ respectively.
Now, by using the condition for the two planes intersection where $\lambda $is any constant as:
${{P}_{2}}+\lambda {{P}_{1}}=0$
So, by substituting the values of both the planes in above equation, we get:
$\begin{align}
  & 2x+3y+4z-5+\lambda \left( x+y+z-1 \right)=0 \\
 & \Rightarrow \left( 2+\lambda \right)x+\left( 3+\lambda \right)y+\left( 4+\lambda \right)z-\left( 5+\lambda \right)=0 \\
\end{align}$
Now, we know that for the above stated plane to be perpendicular to the plane $x-y+z=0$, the dot product of the normal vectors of the planes be zero.
So, by using the above condition, we get:
$\left( 2+\lambda \right)\left( 1 \right)+\left( 3+\lambda \right)\left( -1 \right)+\left( 4+\lambda \right)\left( 1 \right)=0$
So, by solving the above expression, we get the value of $\lambda $as:
$\begin{align}
  & \left( 2+\lambda \right)-\left( 3+\lambda \right)+\left( 4+\lambda \right)=0 \\
 & \Rightarrow 3+\lambda =0 \\
 & \Rightarrow \lambda =-3 \\
\end{align}$
Then, by substituting the value of $\lambda $ as -3 in the equation of the plane, we get:
$\begin{align}
  & \left( 2+\left( -3 \right) \right)x+\left( 3+\left( -3 \right) \right)y+\left( 4+\left( -3 \right) \right)z-\left( 5+\left( -3 \right) \right)=0 \\
 & \Rightarrow -x+z-2=0 \\
 & \Rightarrow x-z+2=0 \\
\end{align}$
So, the equation of the required plane is $x-z+2=0$.
Then, to get the distance d of the plane from the origin where A, B, C and D are the coefficients of x, y, z and constant term in plane equation and $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ is the point to which value is calculated is given by:
$d=\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+C{{z}_{1}}-D}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}} \right|$
So, we need the value from origin which means ${{x}_{1}}={{y}_{1}}={{z}_{1}}=0$.
Then, we get the distance d as:
$\begin{align}
  & d=\left| \dfrac{-2}{\sqrt{{{1}^{2}}+{{0}^{2}}+{{\left( -1 \right)}^{2}}}} \right| \\
 & \Rightarrow d=\left| \dfrac{-2}{\sqrt{1+0+1}} \right| \\
 & \Rightarrow d=\left| \dfrac{-2}{\sqrt{2}} \right| \\
 & \Rightarrow d=\sqrt{2} \\
\end{align}$
So, we get the distance of the plane from the origin as $\sqrt{2}$units.

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.