
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
163.5k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

What is Normality in Chemistry?

Chemistry Electronic Configuration of D Block Elements: JEE Main 2025

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
