Question

If the planes \[3x - 2y + 2z + 17 = 0\] and \[4x + 3y - kz = 25\] are mutually perpendicular, then what is the value of \[k\]?
A. 3
B. \[ - 3\]
C. 9
D. \[ - 6\]

Answer 1

Hint: Here, the equations of two mutually perpendicular planes are given. Apply the condition required for the two planes to be perpendicular to each other on the given equations of the plane. Solve it and get the required answer.

Formula used: If two planes \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] are perpendicular to each other then, \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\].

Complete step by step solution: Given: The planes \[3x - 2y + 2z + 17 = 0\] and \[4x + 3y - kz = 25\] be mutually perpendicular to each other.

In the given planes, \[\left( {3, - 2,2} \right)\] are the direction ratios of the plane \[3x - 2y + 2z + 17 = 0\] and \[\left( {4,3, - k} \right)\] are the direction ratios of the plane \[4x + 3y - kz - 25 = 0\].

Apply the condition required for the perpendicular planes, if two planes \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] are perpendicular to each other then, \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\].
From, the given information, we get
\[{a_1} = 3,{b_1} = - 2,{c_1} = 2\] and \[{a_2} = 4,{b_2} = 3,{c_2} = - k\]
So, for the given equations of the plane
\[\left( 3 \right)\left( 4 \right) + \left( { - 2} \right)\left( 3 \right) + \left( 2 \right)\left( { - k} \right) = 0\]
\[ \Rightarrow 12 - 6 - 2k = 0\]
\[ \Rightarrow 6 - 2k = 0\]
\[ \Rightarrow 2k = 6\]
\[ \Rightarrow k = 3\]

Thus, Option (A) is correct.

Note: The condition required for the perpendicular planes \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\] is obtained from the formula of the angle between the two planes.
The angle between the plane \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] is \[\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\].
When the planes are perpendicular, the angle between them is \[{90^ \circ }\].
We know that \[\cos \left( {{{90}^ \circ }} \right) = 0\]. So, the numerator must be 0.
Thus, \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\].