
The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is \[\left( {2,4, - 3} \right)\]. Then find the equation of the plane.
A. \[2x - 4y - 3z = 29\]
B. \[2x - 4y + 3z = 29\]
C. \[2x + 4y - 3z = 29\]
D. None of these
Answer
162k+ views
Hint: First, calculate the general equation of the plane on the basis of the given point. Then, calculate the direction ratios of the plane. After that, substitute the values of the direction ratios in the equation and solve it to get the required answer.
Formula used: The equation of plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\] is: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\] .
Complete step by step solution: Given: The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is \[\left( {2,4, - 3} \right)\].
Let consider, \[\left( {a,b,c} \right)\] are the direction ratios of the required plane.
So, the equation of the plane passing through the point \[\left( {2,4, - 3} \right)\] is:
\[a\left( {x - 2} \right) + b\left( {y - 4} \right) + c\left( {z + 3} \right) = 0\] \[.....\left( 1 \right)\]
Since the line joining the origin and the point \[\left( {2,4, - 3} \right)\] is normal to the plane.
So, the direction ratios of normal to the plane are \[\left( {2,4, - 3} \right)\].
We get, \[a = 2,b = 4,c = - 3\]
Now substitute these values in the equation \[\left( 1 \right)\].
\[2\left( {x - 2} \right) + 4\left( {y - 4} \right) - 3\left( {z + 3} \right) = 0\]
\[ \Rightarrow 2x - 4 + 4y - 16 - 3z - 9 = 0\]
\[ \Rightarrow 2x + 4y - 3z - 29 = 0\]
\[ \Rightarrow 2x + 4y - 3z = 29\]
Thus, the equation of the required plane is \[2x + 4y - 3z = 29\].
Thus, Option (C) is correct.
Note: Students often get confused between the direction ratios and direction cosines. Remember the following formulas:
For the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\]:
Direction ratios: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]
Direction cosines: \[\left( {\dfrac{{{x_2} - {x_1}}}{w},\dfrac{{{y_2} - {y_1}}}{w},\dfrac{{{z_2} - {z_1}}}{w}} \right)\], where \[w = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]
Formula used: The equation of plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\] is: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\] .
Complete step by step solution: Given: The co-ordinates of the foot of the perpendicular drawn from the origin to a plane is \[\left( {2,4, - 3} \right)\].
Let consider, \[\left( {a,b,c} \right)\] are the direction ratios of the required plane.
So, the equation of the plane passing through the point \[\left( {2,4, - 3} \right)\] is:
\[a\left( {x - 2} \right) + b\left( {y - 4} \right) + c\left( {z + 3} \right) = 0\] \[.....\left( 1 \right)\]
Since the line joining the origin and the point \[\left( {2,4, - 3} \right)\] is normal to the plane.
So, the direction ratios of normal to the plane are \[\left( {2,4, - 3} \right)\].
We get, \[a = 2,b = 4,c = - 3\]
Now substitute these values in the equation \[\left( 1 \right)\].
\[2\left( {x - 2} \right) + 4\left( {y - 4} \right) - 3\left( {z + 3} \right) = 0\]
\[ \Rightarrow 2x - 4 + 4y - 16 - 3z - 9 = 0\]
\[ \Rightarrow 2x + 4y - 3z - 29 = 0\]
\[ \Rightarrow 2x + 4y - 3z = 29\]
Thus, the equation of the required plane is \[2x + 4y - 3z = 29\].
Thus, Option (C) is correct.
Note: Students often get confused between the direction ratios and direction cosines. Remember the following formulas:
For the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\]:
Direction ratios: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]
Direction cosines: \[\left( {\dfrac{{{x_2} - {x_1}}}{w},\dfrac{{{y_2} - {y_1}}}{w},\dfrac{{{z_2} - {z_1}}}{w}} \right)\], where \[w = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
