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Perpendicular Distance Of A Point From A Plane - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Distance of a Point From a Plane is considered one the most difficult concept.

  • 24 Questions around this concept.

Solve by difficulty

QuestionEASY

 The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular

to both the lines \frac{x-1}{1}= \frac{y+2}{-2}= \frac{z-4}{3}  and \frac{x-2}{2}= \frac{y+1}{-1}= \frac{z+7}{-1},is:
 

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If the points (1, 1, \lambda) and (-3, 0, 1) are equidistant from the plane, 3x+4y-12z+13=0, then \lambdasatisfies the equation :

View Solution

Distance between two parallel planes 2x+y+2z=8\: and \: 4x+2y+4z+5=0 is :

View Solution

 The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and  2x −2y + z + 12=0, is :

 

View Solution

Concepts Covered - 1

Distance of a Point From a Plane

The perpendicular distance (D) from a point having position vector \vec{\mathbf a} to the plane \vec{\mathbf r}\cdot \vec{\mathbf n}=d is given by

\mathbf D=\frac{|\vec{\mathbf a} \cdot \overrightarrow{\mathbf{n}}-d|}{|\overrightarrow{\mathbf{n}}|}

Consider a point P with position vector \vec{\mathbf a} and a plane π1 whose equation is \vec{\mathbf r}\cdot \vec{\mathbf n}=d 

Let R be a point in the plane such that \overrightarrow{PR} is orthogonal to the plane π1.

since, line PR passes through P(a) and is parallel to the vector \vec{\mathbf n} which is normal to the plane π1. So, vector equation of line PR is 

\vec{\mathbf r}=\vec{\mathbf a}+\lambda \vec{\mathbf n}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots\text{(i)}

Point R is the intersection of Eq. (i) and the given plane π1.

\\\therefore \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left (\vec{\mathbf a}+\lambda \vec{\mathbf n} \right )\cdot \vec{\mathbf n}=d\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vec{\mathbf a}\cdot \vec{\mathbf n}+\lambda \;\vec{\mathbf n}\cdot\vec{\mathbf n}=d\\\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \lambda=\frac{d-\vec{\mathbf a}\cdot\vec{\mathbf n}}{\left | \vec{\mathbf n} \right |^2}

On putting the value of λ in Eq. (i), we obtain the position vector of R given by

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec {\mathbf r}=\vec{\mathbf a}+\left ( \frac{d-\vec{\mathbf a}\cdot\vec{\mathbf n}}{\left | \vec{\mathbf n} \right |^2} \right )\vec{\mathbf n}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{\mathbf{P R}}=\text { Position vector of } R-\text { Position vector of } P\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\vec{\mathbf a}+\left ( \frac{d-\vec{\mathbf a}\cdot\vec{\mathbf n}}{\left | \vec{\mathbf n} \right |^2} \right )\vec{\mathbf n}-\vec{\mathbf a}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\overrightarrow{\mathbf{P R}}=\left ( \frac{d-\vec{\mathbf a}\cdot\vec{\mathbf n}}{\left | \vec{\mathbf n} \right |^2} \right )\vec{\mathbf n}\\\Rightarrow \mathrm{\;\;\;\;\;\;}\left | \overrightarrow{\mathbf{P R}} \right |=\left | \left ( \frac{d-\vec{\mathbf a}\cdot\vec{\mathbf n}}{\left | \vec{\mathbf n} \right |^2} \right )\vec{\mathbf n} \right |

\\\Rightarrow \mathrm{\;\;\;\;\;\;}\left | \overrightarrow{\mathbf{P R}} \right |= \frac{\left | d-\left ( \vec{\mathbf a}\cdot\vec{\mathbf n} \right ) \right |}{\left | \vec{\mathbf n} \right |}\\ \mathrm{or\;\;\;\;\;\;\;\;\;\;\;\;\;}D= \frac{\left | \left ( \vec{\mathbf a}\cdot\vec{\mathbf n} \right ) -d\right |}{\left | \vec{\mathbf n} \right |}


Cartesian Form

Let P(x1, y1, z1) be the given point with position vector \vec{\mathbf a} and ax + by + cz + d = 0 be the Cartesian equation of the given plane. Then

\\ \mathrm{\;\;}\vec{ a} =x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k} \\ \overrightarrow =\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}

Hence, from Vector form of the perpendicular from P to the plane is

\\\left|\frac{\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{k}\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})-(-d)}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

 

Distance Between The Parallel Planes

The distance between the two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by 
D=\left|\frac{\left(d_{2}-d_{1}\right)}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Let P(x1, y1, z1) be any point in the plane  ax + by + cz + d1 = 0.

Then the distance of the point P from plane  ax + by + cz + d2 = 0 is

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}D=\left|\frac{ax_1+by_1+cz_1+d_2}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\\\\\text{Also,}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;ax_1+by1+cz_1+d_1=0\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathbf{D=\left|\frac{\left(d_{2}-d_{1}\right)}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|}  

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Distance of a Point From a Plane

Mathematics for Joint Entrance Examination JEE (Advanced) : Vectors and 3D Geometry

Page No. : 4.43

Line : 22

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