Question

If the planes \[x + 2y + kz = 0\] and \[2x + y - 2z = 0\] are at right angles. Then what is the value of \[k\]?
A. \[ - \dfrac{1}{2}\]
B. \[\dfrac{1}{2}\]
C. \[ - 2\]
D. 2

Answer 1

Hint: Here, the equations of the two perpendicular planes are given. First, apply the condition required for the two planes to be perpendicular. Then, substitute the values in the equation of dot product of the normal vectors of the planes. In the end, solve the equation and get the required answer.

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

Complete step by step solution: Given:
The planes \[x + 2y + kz = 0\] and \[2x + y - 2z = 0\] are at right angles.

We know that if two planes \[{a_1}x + {b_1}y + {c_1}z = {d_1}\] and \[{a_2}x + {b_2}y + {c_2}z = {d_2}\] are perpendicular, then \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\].
It is given that the planes \[x + 2y + kz = 0\] and \[2x + y - 2z = 0\] are at right angles.
Comparing the equation of the planes, we get
\[{a_1} = 1,{b_1} = 2,{c_1} = k\] and \[{a_2} = 2,{b_2} = 1,{c_2} = - 2\]
So, apply the above condition of the perpendicular planes on the given planes.
We get,
\[\left( 1 \right)\left( 2 \right) + \left( 2 \right)\left( 1 \right) + k\left( { - 2} \right) = 0\]
\[ \Rightarrow 2 + 2 - 2k = 0\]
\[ \Rightarrow 4 - 2k = 0\]
\[ \Rightarrow 2k = 4\]
\[ \Rightarrow k = 2\]
Thus, the required value is \[k = 2\].

Thus, Option (D) is correct.

Note: The required condition for the two planes to be perpendicular to each other is obtained from the formula of the angle between the two planes \[\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\]. When two planes are perpendicular, then \[\cos {90^ \circ } = 0\]. Since, the value of the denominator cannot be 0. So, value of the numerator must be 0.