Question

The plane through the line of intersection of the planes \[x-y+2=0,y-2z-1=0\] whose distance from the origin is unity.
\[\begin{align}
  & A.\text{ }x-2y+2z+3=0 \\
 & B.\text{ }2x-y+2z+3=0 \\
 & C.\text{ }x+2y+2z+3=0 \\
 & D.\text{ }2x+y+2z+3=0 \\
\end{align}\]

Answer 1

Hint: At first find the intersection of planes by using formula, if planes are given such as,
\[ax\text{ + by + cz + d = 0}\] and \[ex\text{ + fy + gz + h = 0}\] then their intersection would be,
\[ax\text{ + by + cz + d + }\lambda \text{ }\left( ex\text{ + fy + gz + h} \right)\text{ = 0}\] if $\lambda $ is a real number.
After this apply the formula of finding distance between point and plane given that point is (a, b, c) and if equation of plane is \[ax\text{ + by + cz + d = 0}\] so we get,
\[\dfrac{\left| Aa\text{ + Bb + Cc + D} \right|}{\sqrt{{{A}^{2}}\text{ + }{{\text{B}}^{2}}\text{ + }{{\text{C}}^{2}}}}\] Then, equate it with 1 to find values of $\lambda $ so as to find equations of plane.

Complete step-by-step answer:
In the question we are said that there is a plane which passes through the intersection of the planes \[x\text{ - y + z = 0}\] and \[y\text{ - 2z - 1 = 0}\] also the distance from the origin is unity.
So the planes given in the question are \[x\text{ - y + z = 0}\] and \[y\text{ - 2z - 1 = 0}\]
If two planes are given let’s say,
\[ax\text{ + by + cz + d = 0}\] and \[ex\text{ + fy + gz + h = 0}\]
Then their intersection can be represented using formula,
\[ax\text{ + by + cz + d + }\lambda \text{ }\left( ex\text{ + fy + gz + h} \right)\text{ = 0}\]
Where $\lambda $ represents any real number.
So, we can write intersection as,
\[x\text{ + }\left( \lambda \text{ - 1} \right)y\text{ + }\left( -2\lambda \right)z\text{ + }\left( 2\text{ - }\lambda \right)\text{ = 0}\]
Where $\lambda $ represents any real number.
Now we are given that distance of plane from origin is 1 unit.
So here we will use formula that if a point (e, f, g) is given and a plane \[Ax\text{ + By + Cz + D = 0}\] is given then the distance between them will be,
\[\dfrac{\left| \left( Ae\text{ + Bf + Cg + D} \right) \right|}{\sqrt{{{A}^{2}}\text{ + }{{\text{B}}^{2}}\text{ + }{{\text{C}}^{2}}}}=\text{Distance}\]
So here point is (O, O, O) and the equation of plane is \[x\text{ + }\left( \lambda \text{ - 1} \right)y\text{ + }\left( -2\lambda \right)z\text{ + }\left( 2\text{ - }\lambda \right)\text{ = 0}\] and distance between them is 1 so,
\[\dfrac{\left| 1.0\text{ + }\left( \lambda -1 \right).0\text{ + }\left( -2\lambda \right).0\text{ + }\left( 2-\lambda \right) \right|}{\sqrt{{{\left( 1 \right)}^{2}}\text{ + }{{\left( \lambda -1 \right)}^{2}}\text{ + }{{\left( -2\lambda \right)}^{2}}}}\text{ = 1}\]
Hence on simplification and cross multiplication we get,
\[\left| 2-\lambda \right|\text{ = }\sqrt{1\text{ + }{{\left( \lambda -1 \right)}^{2}}\text{ + 4}{{\lambda }^{2}}}\]
Now on squaring both the sides we get,
\[{{\left( 2\text{ - }\lambda \right)}^{2}}\text{ = 1 + }{{\left( \lambda \text{ - 1} \right)}^{2}}\text{ + 4}{{\lambda }^{2}}.\]
Now expanding by using the formula \[{{\left( a\text{ - b} \right)}^{2}}\text{ = }{{\text{a}}^{2}}\text{ + }{{\text{b}}^{2}}\text{ - 2ab}\] we get
\[\begin{align}
  & 4\text{ - 4}\lambda \text{ + }{{\lambda }^{2}}\text{ = 1 + }{{\lambda }^{2}}\text{ - 2}\lambda \text{ + 1 + 4}{{\lambda }^{2}} \\
 & \Rightarrow {{\lambda }^{2}}\text{-4}\lambda +4\text{ = 5}{{\lambda }^{2}}-2\lambda +2 \\
\end{align}\]
Now subtracting \[\left( {{\lambda }^{2}}\text{-4}\lambda +4 \right)\text{ }\] to both the sides we get,
\[\begin{align}
  & \left( \text{5}{{\lambda }^{2}}-2\lambda +2 \right)-\left( {{\lambda }^{2}}-4\lambda +4 \right)=0 \\
 & \Rightarrow 4{{\lambda }^{2}}+2\lambda -2=0 \\
\end{align}\]
Now dividing by 2 throughout we get,
\[2{{\lambda }^{2}}\text{ + }\lambda \text{ - 1 = 0}\]
We can also write the same expression as,
\[2{{\lambda }^{2}}\text{ - }\lambda \text{ + 2}\lambda \text{ - 1 = 0}\]
Which can be written as,
\[\lambda \text{ }\left( 2\lambda \text{ - 1} \right)\text{ + 1}\left( 2\lambda \text{ - 1} \right)\text{ = 0}\]
Which can be factorized as,
\[\left( \lambda \text{ + 1} \right)\left( 2\lambda \text{ - 1} \right)\text{ = 0}\]
So values for which $\lambda $ satisfies is $\lambda =\text{ -1}$ and $\lambda \text{ = }\dfrac{1}{2}$
The plane of intersection was \[x\text{ + }\left( \lambda \text{ - 1} \right)y\text{ + }\left( -2\lambda \right)z\text{ + }\left( 2-\lambda \right)\text{ = 0}\]
For $\lambda =\text{ -1,}$ the plane equation will be,
\[\begin{align}
  & x\text{ + }\left( -1-1 \right)y\text{ + }\left( -2\times -1 \right)z\text{ + }\left( 2-\left( -1 \right) \right)\text{ = 0} \\
 & \Rightarrow \text{x - 2y + 2z + 3 = 0} \\
\end{align}\]
and for $\lambda \text{ = }\dfrac{1}{2},$ the plane equation will be
\[\begin{align}
  & x\text{ + }\left( \dfrac{1}{2}-1 \right)y\text{ + }\left( -2\times \dfrac{1}{2} \right)z\text{ + }\left( 2-\dfrac{1}{2} \right)\text{ = 0} \\
 & \Rightarrow \text{x + }\left( \dfrac{-y}{2} \right)-z\text{ + }\dfrac{3}{2}\text{ = 0} \\
\end{align}\]
Now by multiplying by 2 throughout we get,
\[2x\text{ - y - 2z + 3 = 0}\]
Hence the equations of planes are
\[x\text{ - 2y + 2z + 3 = 0 and 2x -y - 2z + 3 = 0}\]

So, the correct answers are “Option A and B”.

Note: After finding the value of $\lambda $ from the equation, one should check for their own purpose by again substituting it, before substituting it in the main equation of planes. The students must note that here we have two correct options for each value of $\lambda $ obtained. So, while choosing options, they must make sure to mark both A and B as right options. Also, since we have been given that distance from origin is 1, the numerator in the formula \[\dfrac{\left| Aa\text{ + Bb + Cc + D} \right|}{\sqrt{{{A}^{2}}\text{ + }{{\text{B}}^{2}}\text{ + }{{\text{C}}^{2}}}}\] can be reduced to just D. So this point can be used to use a simpler formula as \[\dfrac{\left| D \right|}{\sqrt{{{A}^{2}}\text{ + }{{\text{B}}^{2}}\text{ + }{{\text{C}}^{2}}}}\] .