Question

Find the equation of plane containing the line $2x - y + z - 3 = 0$; $3x + y + z = 5$ and at a distance of $\dfrac{1}{{\sqrt 6 }}$ from the point $\left( {2,1, - 1} \right)$.
A. $2x - y + z - 3 = 0$
B. $62x + 29y + 19z - 105 = 0$
C. $3x + y + z = 5$
D. None of these

Answer 1

In this problem, first we need to find the equation of the plane containing the given two lines. Next, find the distance of the plain from the given points and hence find the equations of the planes.

Complete step-by-step answer:
The equation of the plane containing the lines $2x - y + z - 3 = 0$ and $3x + y + z = 5$ is as follows:

$$\,\,\,\,\,\,2x - y + z - 3 + \lambda \left( {3x + y + z - 5} \right) = 0 \\\ \Rightarrow 2x - y + z - 3 + 3x\lambda + y\lambda + z\lambda - 5\lambda = 0 \\\ \Rightarrow \left( {2 + 3\lambda } \right)x + \left( {\lambda - 1} \right)y + \left( {\lambda + 1} \right)z - 3 - 5\lambda = 0 \\\$$

Since, the distance of the plane $\left( {2 + 3\lambda } \right)x + \left( {\lambda - 1} \right)y + \left( {\lambda + 1} \right)z - 3 - 5\lambda = 0$ from the points $\left( {2,1, - 1} \right)$ is$\dfrac{1}{{\sqrt 6 }}$, it can be written as follows:

$$\,\,\,\,\,\left| {\dfrac{{2\left( {2 + 3\lambda } \right) + 1\left( {\lambda - 1} \right) - 1\left( {\lambda + 1} \right) - 3 - 5\lambda }}{{\sqrt {{{\left( {2 + 3\lambda } \right)}^2} + {{\left( {\lambda - 1} \right)}^2} + {{\left( {\lambda + 1} \right)}^2}} }}} \right| = \dfrac{1}{{\sqrt 6 }} \\\ \Rightarrow \left| {\dfrac{{4 + 6\lambda + \lambda - 1 - \lambda - 1 - 3 - 5\lambda }}{{\sqrt {4 + 9{\lambda ^2} + 12\lambda + 1 + {\lambda ^2} - 2\lambda + 1 + {\lambda ^2} + 2\lambda } }}} \right| = \dfrac{1}{{\sqrt 6 }} \\\ \Rightarrow \left| {\dfrac{{\lambda - 1}}{{\sqrt {11{\lambda ^2} + 12\lambda + 6} }}} \right| = \dfrac{1}{{\sqrt 6 }} \\\ \Rightarrow 6{\left( {\lambda - 1} \right)^2} = 11{\lambda ^2} + 12\lambda + 6 \\\ \Rightarrow 6\left( {1 + {\lambda ^2} - 2\lambda } \right) = 11{\lambda ^2} + 12\lambda + 6 \\\$$

Further, simplify the above equation.

$$\,\,\,\,\,6 + 6{\lambda ^2} - 12\lambda = 11{\lambda ^2} + 12\lambda + 6 \\\ \Rightarrow 5{\lambda ^2} + 24\lambda = 0 \\\ \Rightarrow \lambda \left( {5\lambda + 24} \right) = 0 \\\ \Rightarrow \lambda = 0\,\,{\text{or}}\,\,\dfrac{{ - 24}}{5} \\\$$

Now, substitute 0 for $\lambda$ in equation of plane$\left( {2 + 3\lambda } \right)x + \left( {\lambda - 1} \right)y + \left( {\lambda + 1} \right)z - 3 - 5\lambda = 0$.

$$\,\,\,\,\,\left( {2 + 3\left( 0 \right)} \right)x + \left( {\left( 0 \right) - 1} \right)y + \left( {\left( 0 \right) + 1} \right)z - 3 - 5\left( 0 \right) = 0 \\\ \Rightarrow 2x - y + z - 3 = 0 \\\$$

Further, substitute $- \dfrac{{24}}{5}$ for $\lambda$ in equation of plane$\left( {2 + 3\lambda } \right)x + \left( {\lambda - 1} \right)y + \left( {\lambda + 1} \right)z - 3 - 5\lambda = 0$.

$$\,\,\,\,\,\left( {2 + 3\left( { - \dfrac{{24}}{5}} \right)} \right)x + \left( {\left( { - \dfrac{{24}}{5}} \right) - 1} \right)y + \left( {\left( { - \dfrac{{24}}{5}} \right) + 1} \right)z - 3 - 5\left( { - \dfrac{{24}}{5}} \right) = 0 \\\ \Rightarrow - \dfrac{{62}}{5}x - \dfrac{{29}}{5}y - \dfrac{{19}}{5}z - 3 + 24 = 0 \\\ \Rightarrow - \dfrac{{62}}{5}x - \dfrac{{29}}{5}y - \dfrac{{19}}{5}z + 21 = 0 \\\ \Rightarrow - 62x - 29y - 19z + 105 = 0 \\\ \Rightarrow 62x + 29y + 19z - 105 = 0 \\\$$

Since, the equation of the planes are $2x - y + z - 3 = 0$
and$62x + 29y + 19z - 105 = 0$
, therefore, option A and B are correct.

Note: The equation of the plane containing two lines ${L_1}$ and ${L_2}$ is${L_1} + \lambda {L_2} = 0$. There are two planes containing the same lines and at the same distance from the given point.