Question

In the space the equation \[by + cz + d = 0\]represents a plane perpendicular to the plane:
a) YOZ
b) z = k
c) ZOX
d) XOY

Answer 1

Hint: Determine the normal vector of the given plane as well as the normal vectors of the planes in the options. Find the dot product between the normal vectors in each case. If the dot product is zero, then the corresponding plane is perpendicular to the given plane.



Formula Used:The normal vector to the plane \[ax + by + cz + d = 0\]is \[\overrightarrow n = a\widehat i + b\widehat j + c\widehat k\].
Equation of YOZ plane is x = 0.
Equation of ZOX plane is y = 0.
Equation of XOY plane is z = 0.
Two vectors \[\overrightarrow a \] and \[\overrightarrow b \]are perpendicular, then \[\overrightarrow a \bullet \overrightarrow b = 0\]
\[\widehat i \bullet \widehat j = \widehat j \bullet \widehat k = \widehat k \bullet \widehat i = 0\]
\[\widehat i \bullet \widehat i = \widehat j \bullet \widehat j = \widehat k \bullet \widehat k = 1\]



Complete step by step solution:The normal vector to the plane \[by + cz + d = 0\]is \[{\overrightarrow n _{_1}} = b\widehat j + c\widehat k\].
Choice (a):
Equation of YOZ plane is x = 0.
The normal vector to the plane x = 0 is \[{\overrightarrow n _{_2}} = \widehat i\].
Dot product of the normal vectors = \[{\overrightarrow n _{_1}} \bullet {\overrightarrow n _{_2}}\]
                                                        = \[\left( {b\widehat j + c\widehat k} \right) \bullet \widehat i\]
                                                        = 0
\[ \Rightarrow \]\[by + cz + d = 0\] is perpendicular to the plane YOZ.
Choice (b):
Equation of the plane is z = k.
The normal vector to the plane z = k is \[{\overrightarrow n _{_3}} = \widehat k\].
Dot product of the normal vectors = \[{\overrightarrow n _{_1}} \bullet {\overrightarrow n _{_3}}\]
                                                        = \[\left( {b\widehat j + c\widehat k} \right) \bullet \widehat k\]
                                                        = \[c \ne 0\]
\[ \Rightarrow \]\[by + cz + d = 0\] is not perpendicular to the plane z = k.
Choice (c):
Equation of ZOX plane is y = 0.
The normal vector to the plane y = 0 is \[{\overrightarrow n _{_4}} = \widehat j\].
Dot product of the normal vectors = \[{\overrightarrow n _{_1}} \bullet {\overrightarrow n _4}\]
                                                        = \[\left( {b\widehat j + c\widehat k} \right) \bullet \widehat j\]
                                                        = \[b \ne 0\]
\[ \Rightarrow \]\[by + cz + d = 0\] is not perpendicular to the plane ZOX.
Choice (d):
Equation of XOY plane is z = 0.
The normal vector to the plane z = 0 is \[{\overrightarrow n _{_5}} = \widehat k\].
Dot product of the normal vectors = \[{\overrightarrow n _{_1}} \bullet {\overrightarrow n _5}\]
                                                        = \[\left( {b\widehat j + c\widehat k} \right) \bullet \widehat k\]
                                                        = \[c \ne 0\]
\[ \Rightarrow \]\[by + cz + d = 0\] is not perpendicular to the plane XOY.



Option ‘a’ is correct



Note: Student may find difficulty in writing the normal vectors for each planes in the choices.
Student may wrongly assumes that \[\widehat i \bullet \widehat j = \widehat j \bullet \widehat k = \widehat k \bullet \widehat i = 1\] and \[\widehat i \bullet \widehat i = \widehat j \bullet \widehat j = \widehat k \bullet \widehat k = 0\]