Question

Find the value of \[k\] for which the planes \[3x - 6y - 2z = 7\] and \[2x + y - kz = 5\] are perpendicular to each other.
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Here, the equations of two planes are given. Apply the condition for two planes to be perpendicular and solve it to get the required value of \[k\].



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 equations of the planes are \[3x - 6y - 2z = 7\] and \[2x + y - kz = 5\].

Now compare the equations of planes with \[{a_1}x + {b_1}y + {c_1}z + {d_1} = 0\] and \[{a_2}x + {b_2}y + {c_2}z + {d_2} = 0\] respectively.
We get,
\[{a_1} = 3,{b_1} = - 6,{c_1} = - 2\] and \[{a_2} = 2,{b_2} = 1,{c_2} = - k\]
We know that, if the planes are perpendicular to each other, then \[{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\].
Substitute the values in the above condition.
Then,
\[\left( 3 \right)\left( 2 \right) + \left( { - 6} \right)\left( 1 \right) + \left( { - 2} \right)\left( { - k} \right) = 0\]
\[ \Rightarrow 6 + - 6 + 2k = 0\]
\[ \Rightarrow 2k = 0\]
\[ \Rightarrow k = 0\]



Option ‘A’ is correct



Note: If the two planes are perpendicular, then their normal vectors must be perpendicular. So, the dot product of the normal vectors of both planes is 0.