# Examples on Equations of Three Dimensional Planes Set 3

**Example – 15**

Find the equation of the plane (s) bisecting the angle(s) between two given planes

\[\begin{align} {P_1} \equiv {a_1}x + {b_1}x + {c_1}z + {d_1} = 0 \\\\ {P_2} \equiv {a_2}x + {b_2}y + {c_2}z + {d_2} = 0 \\ \end{align} \]

**Solution:** Note that as in the case of the intersection of straight lines, there will be two (supplementary) angles formed when two planes intersect: one will be acute and the other obtuse (or both could be right). The angle bisector plane of two planes has essentially the same property as the angle bisector of two lines: any point on the angle bisector plane of the planes *P*_{1} and *P*_{2} will be equidistant from *P*_{1} and *P*_{2}.

If we assume an arbitrary point *S*(*x*, *y*, *z*) on the angle bisector plane(s) of *P*_{1} and *P*_{2} , we have,

Distance of *S* from *P*_{1} = Distance of *S *from *P*_{2}

\[\begin{align} \Rightarrow \quad \frac{{\left| {{a_1}x + {b_1}y + {c_1}z + {d_1}} \right|}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} = \frac{{\left| {{a_2}x + {b_2}y + {c_2}z + {d_2}} \right|}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }} \end{align}\]

\[\begin{align}\Rightarrow \quad \boxed{\frac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} = \pm \frac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \\ \end{align} \]

As expected, we get two angle bisector planes, one corresponding to the “+” and one to the “–” sign.

As in the case of straight line angle bisectors, we can prove that the equation of the angle bisector containing the origin will be given by the “+” sign if *d*_{1} and *d*_{2} are of the same sign. You are urged to prove this as an exercise.

For two planes with equations

\[\begin{align} {P_1} \equiv {a_1}x + {b_1}y + {c_1}z + {d_1} = 0 & \equiv & \vec r \cdot {{\vec n}_1} + {d_1} = 0 \\\\ {P_2} \equiv {a_2}x + {b_2}y + {c_2}z + {d_2} = 0 & \equiv & \vec r \cdot {{\vec n}_2} + {d_2} = 0 \\ \end{align} \]

we have already proved in the chapter on vectors that **any **plane passing through the intersection line of *P*_{1} and *P*_{2} can be written as

\[{{P}_{1}}+\lambda {{P}_{2}}=0,\lambda \in \mathbb{R}\]

Let us use this to solve a problem.

**Example – 16**

Find the equation of the plane passing through the line of intersection of

\[\begin{align} {P_1} \equiv x + 3y - 6 = 0\\\\ {P_2} \equiv 3x - y + 4z = 0 \\ \end{align} \]

and at a unit distance from the origin.

**Solution:** Any plane through the intersection line of *P*_{1} and *P*_{2 }can be written as

\[\begin{align}&\qquad\;\; {P_1} + \lambda {P_2} = 0 \\\\ &\Rightarrow \quad \left( {1 + 3\lambda } \right)x + \left( {3 - \lambda } \right)y + 4\lambda z - 6 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ \end{align} \]

The distance of this plane from the origin (0, 0, 0) is 1. We thus have, using the formula for the distance of a point from a plane,

\[\begin{align} &\qquad \frac{{\left| {\left( {1 + 3\lambda } \right)0 + \left( {3 - \lambda } \right)0 + 4\lambda \left( 0 \right) - 6} \right|}}{{\sqrt {{{\left( {1 + 3\lambda } \right)}^2} + {{\left( {3 - \lambda } \right)}^2} + {{\left( {4\lambda } \right)}^2}} }} = 1 \\\\ &\Rightarrow \quad {\left( {1 + 3\lambda } \right)^2} + {\left( {3 - \lambda } \right)^2} + {\left( {4\lambda } \right)^2} = 36\\\\ &\Rightarrow \quad \lambda = \pm 1 \\ \end{align} \]

Thus, if fact two such planes will exist . Using the values of obtained in (1), the equations of these two planes will be \(2x + y + 2z + 3 = 0\) and \( - x + 2y - 2z + 3 = 0.\)