Question

A plane which bisects the angle between the two given planes $ ~2x-y+2z-4=0$ and $ x+2y+2z-2=0 $\[\]
A.$\left( 2,4,1 \right)$\[\]
B. $\left( 2,-4,1 \right)$\[\]
C. $\left( 1,4,-1 \right)$\[\]
D. $\left( 1,-4,1 \right)$\[\]

Answer 1

Hint: Take a point $P$ on the angle bisector plane. Use the plane ($ax+by+cz=d$) to point $P\left( {{x}_{o}},{{y}_{0}},{{z}_{0}} \right)$ distance formula $d=\dfrac{a{{x}_{0}}+b{{y}_{0}}+c{{z}_{0}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ to find out the distance from both the given planes. Equate both the distances to get a pair of linear equations. Put each point in both the equations to test whether they satisfy or not. \[\]

Complete step-by-step answer:
If there are two planes $ax+by+cz+d=0$ and $lx+my+nz+r=0$ then if a plane bisects the angle between them then we take a point $P$ on the bisector plane. Lets us draw perpendiculars on both the planes, the length of perpendicular will be equal. Using the formula of a plane from a point we obtain the equations bisecting the acute and obtuse angle between the planes
\[\dfrac{a{{x}_{0}}+b{{y}_{0}}+c{{z}_{0}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\dfrac{l{{x}_{0}}+m{{y}_{0}}+n{{z}_{0}}+r}{\pm \sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}\]

The given equations of planes are
\[\begin{align}
  & ~2x-y+2z-4=0...(1) \\
 & x+2y+2z-2=0...(2) \\
\end{align}\]
Let any point on the plane that bisects the angle between planes (1) and (2) denoted as $P$. Then the point will be at equal distance from both the planes . Using the formula of perpendicular distance of a plane from a point, we calculate the distance of $P$ from plane(1)
\[{{d}_{1}}=\left| \dfrac{2x-y+2z-4}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}}} \right|=\left| \dfrac{2x-y+2z-4}{3} \right|\]
Similarly the distance of $P$ from plane(2) is
\[{{d}_{1}}=\left| \dfrac{x+2y-2z-2}{\sqrt{{{2}^{2}}+{{\left( -2 \right)}^{2}}+{{1}^{2}}}} \right|=\left| \dfrac{x+2y-2z-2}{3} \right|\]
As both the distances are equal,
\[\begin{align}
  & {{d}_{1}}={{d}_{2}} \\
 & \Rightarrow \left| \dfrac{2x-y+2z-4}{3} \right|=\left| \dfrac{x+2y-2z-2}{3} \right| \\
 & \Rightarrow \dfrac{2x-y+2z-4}{3}=\pm \dfrac{x+2y-2z-2}{3} \\
\end{align}\]
Taking the positive sign we get one possible equation of the bisector plane that is $x-3y=2$ and taking negative sign we get another possible equation that is $3x+y+z=0$.\[\]
Now checking each point given in the question whether they lie on one of the plane is \[\]
Checking$\left( 2,4,1 \right)$: the point does not satisfy either possible equation of the bisector plane. \[\]
Checking$\left( 2,-4,1 \right)$: the point satisfies the equation $3x+y+z=0$\[\]
Checking$\left( 1,4,-1 \right)$: the point does not satisfy either possible equation of bisector plane\[\]
Checking$\left( 1,-4,1 \right)$: the point does not satisfy either possible equation of bisector plane\[\]

So, the correct answer is “Option B”.

Note: While solving this problem we should not forget the modulus sign on the distance formula otherwise we shall get only the equation of bisector plan as we know there are two equations of bisector planes one for the obtuse angle and acute angle. If the angle between the bisector plane and the plane is less than then the angle is acute. It is noted that angle between two planes $ax+by+cz+d=0$ and $lx+my+nz+r=0$ is given by $ {\cos ^{ - 1}}\left( {\left| {\dfrac{{al + bm + cn}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{l^2} + {m^2} + {n^2}} }}} \right|} \right)$.