
If for a plane, the intercepts on the coordinate axes are 8, 4, 4. Then find the length of the perpendicular from the origin on to the plane.
A. \[\dfrac{8}{3}\]
B. \[\dfrac{3}{8}\]
C. 3
D. \[\dfrac{4}{3}\]
E. \[\dfrac{4}{5}\]
Answer
216.3k+ views
Hint: Here, the intercepts of the plane on the coordinate axes are given. First, by using the intercepts calculate the equation of the plane. Then, calculate the distance between the origin and the plane to get the required answer.
Formula used: The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
Complete step by step solution: Given:
The intercepts of the plane on the coordinate axes are: 8, 4, 4
Let’s calculate the equation of the plane in the intercept form.
8, 4, 4 are the intercepts of the plane.
So, the equation of the plane is:
\[\dfrac{x}{8} + \dfrac{y}{4} + \dfrac{z}{4} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{y \times 2}}{{4 \times 2}} + \dfrac{{z \times 2}}{{4 \times 2}} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{2y}}{8} + \dfrac{{2z}}{8} = 1\]
\[ \Rightarrow \dfrac{{x + 2y + 2z}}{8} = 1\]
\[ \Rightarrow x + 2y + 2z = 8\]
Now calculate the distance between the plane \[x + 2y + 2z = 8\] and the origin \[\left( {0,0,0} \right)\].
So, use the formula of the smallest distance between a plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
Here, \[\left( {{x_1},{y_1},{z_1}} \right) = \left( {0,0,0} \right)\] and the equation of plane is \[x + 2y + 2z = 8\].
Substitute the values in the formula. We get
\[D = \left| {\dfrac{{\left( 1 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) - 8}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt {1 + 4 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt 9 }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{3}} \right|\]
\[ \Rightarrow D = \dfrac{8}{3}\]
Thus, the distance between the origin and the plane is \[\dfrac{8}{3}\].
Thus, Option (A) is correct.
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of question, they will get only one value or the wrong value. So, always remember to apply the modulus function.
Formula used: The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
Complete step by step solution: Given:
The intercepts of the plane on the coordinate axes are: 8, 4, 4
Let’s calculate the equation of the plane in the intercept form.
8, 4, 4 are the intercepts of the plane.
So, the equation of the plane is:
\[\dfrac{x}{8} + \dfrac{y}{4} + \dfrac{z}{4} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{y \times 2}}{{4 \times 2}} + \dfrac{{z \times 2}}{{4 \times 2}} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{2y}}{8} + \dfrac{{2z}}{8} = 1\]
\[ \Rightarrow \dfrac{{x + 2y + 2z}}{8} = 1\]
\[ \Rightarrow x + 2y + 2z = 8\]
Now calculate the distance between the plane \[x + 2y + 2z = 8\] and the origin \[\left( {0,0,0} \right)\].
So, use the formula of the smallest distance between a plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
Here, \[\left( {{x_1},{y_1},{z_1}} \right) = \left( {0,0,0} \right)\] and the equation of plane is \[x + 2y + 2z = 8\].
Substitute the values in the formula. We get
\[D = \left| {\dfrac{{\left( 1 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) - 8}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt {1 + 4 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt 9 }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{3}} \right|\]
\[ \Rightarrow D = \dfrac{8}{3}\]
Thus, the distance between the origin and the plane is \[\dfrac{8}{3}\].
Thus, Option (A) is correct.
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of question, they will get only one value or the wrong value. So, always remember to apply the modulus function.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

