Perpendicular Distance of a Point from a Plane Formula

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Perpendicular distance of a point to a plane is defined as the shortest distance covered from one point to a plane. It is known as the length of the perpendicular which is drawn from that one point to touch the plane. Perpendicular distance of a point from a plane is calculated by using two forms namely vector form and cartesian form. These are the two forms which will help us in measuring the shortest distance between the point and the plane. 

Key Takeaways: Perpendicular, Point on a Plane, Vector Form, Cartesian Form.


Perpendicular Distance of a Point from a Plane

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The shortest distance covered from one point to a plane is known to be along the line which is perpendicular to the plane, or in other words, is known as the perpendicular distance of that one point from the plane. Hence, if we take a normal vector, suppose for example n, to the plane given, a line which runs parallel to n vector which meets the point M will give us the shortest distance of point M from the plane. 

Graphical Representation of perpendicular distance of a point from a plane

Graphical Representation of perpendicular distance of a point from a plane

If we express the point of intersection, for example, I, of the line touching the point M, and the plane on which it touches normally, the point I is known as the point which is closest to the point M on the plane. The distance between the two points M and I will give us the distance of the point M from the plane. We can calculate the shortest distance of a point from a plane using two different methods known as:

  • The Vector method 
  • The Cartesian Method

The video below explains this:

Coordinate Geometry Detailed Video Explanation:


Vector Form

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For example, let us consider a point B to whom the position vector is given by \(\vec{r}\) and a plane P, which is given by the following equation,

\(\vec{r}. \vec{N} = d\)

where; N is normal to the plane.

Now, consider another plane which is parallel to the first plane and that passes through point B, by this, we will get the equation of the second plane where N is normal to the plane. This equation can be written as

\(\vec{r}-\vec{a}. \vec{N} = 0\) 

or in other easier way,

\(\vec{r} . \vec{N} = \vec{a}.\vec{N}\)

Now let’s consider O as the origin of the coordinates, the distance between the first plane from the origin is written as ON. Similarly, the distance of the other plane from the origin O is denoted as ON’. The distance between these two planes is calculated by ON–ON’. This distance is equal to;

ON−ON′= d′ =|d−a. N^ |

We can also find the perpendicular distance of point B on the plane P’ from P. Therefore, we can conclude that for a plane which is denoted by the following equation;

r\(\to\). N\(\to\) = D\(\to\)

and a point B whose position vector we know, we can calculate the perpendicular distance from a point to the plane with the given formula;

\(d = | \vec{r}.\vec{N} - D| / |\vec{N}|\)

Now if we want to calculate the length of the plane from the origin O, we need to substitute the value 0 in the place of the position vector. This will give us;

d = |D|/ |N|

This is how we calculate the perpendicular distance of a point from a plane in vector form.

Vector Form

Vector Form

Read Also: Different Forms of the Equation of line


Cartesian Form

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Suppose a plane that is denoted by the Cartesian equation,

Px + Qy + Rz = S

Now suppose that a point whose position vector and Cartesian coordinates of position vector are given by,

A (x1, y1, z1)

We can also write the position vector in the form;

a\(\to\) = x1 iˆ+ y1 jˆ + z1

For finding the distance of point B from the plane by using the vector form of the formula, we can find the normal vector to the plane, which is given as,

N =Aiˆ+ Bjˆ+ Ckˆ

When we use the formula, the perpendicular distance of point B from the given plane is written as,

\(d = | \vec{a}.\vec{N} - D| / |\vec{N}|\)

d = |Ax1+By1+Cz1−D / √A2+B2+C2

This is the equation that will give us the perpendicular distance of a given point from its plane by using the Cartesian method.


Things to Remember

  • The shortest distance covered from one point to a plane is known to be along the line which is perpendicular to the plane, or in other words, is known as the perpendicular distance of that one point from the plane. 
  • It is calculated by two different forms namely the vector form, and the cartesian form.
  • Formula for calculating the perpendicular distance of a point from a plane in vector form is d = |D| / |N|
  • Formula for calculating the perpendicular distance of a point from a plane in cartesian form is d = |Ax1+By1+Cz1−D / √A2+B2+C2|.

Also Read:


Previous Year's Questions

  1. If aa is a unit vector, then |a×^i|2+|a×^j|2+|a×^k|2=|a×i^|2+|a×j^|2+|a×k^|2= …….[TS EAMCET 2017]
  2. Let →a,→ba→,b→ & →cc→ be non-coplanar unit vectors equally inclined to one another at an acute angle θθ. Then |[→a→b→c]||[a→b→c→]| in terms of θθ is equal to…..[BITSAT 2017]
  3. Two vectors are given by →A=3^i+^j+3^kA→=3i^+j^+3k^ and →B=3^i+5^j−2^kB→=3i^+5j^−2k^. Find the third vector  ….[JKCET 2007]
  4. A constant power is supplied to a rotating disc. The relationship between the angular velocity ωω of the disc and number of rotations (n) made by the disc is governed by...[CBSE CLASS XII]
  5. From a point A with position vector p(^i+^j+^k)p(i^+j^+k^), AB and AC are drawn perpendicular to the lines ….[JEE Main 2018]
  6. If a|(b+c)a|(b+c) and a|(b−c)a|(b−c) where a,b,c∈Na,b,c∈N then…...[KCET 2006]
  7. The vector equation of the symmetrical form of equation of straight line x−53=y+47=z−62x−53=y+47=z−62 is…...[VITEEE 2018]
  8. If a and b are vectors such that |a+b|=|a−b||a+b|=|a−b| then the angle between a and b is….[KCET 2007]
  9. Let →a=^i+^j+√2^k,→b=b1^i+b2^j+√2^ka→=i^+j^+2k^,b→=b1i^+b2j^+2k^ and →c=5^i+^j+√2^kc→=5i^+j^+2k^ be three vectors such that the projection vector of →bb→ on →aa→ , If →a+→ba→+b→ is perpendicular to →cc→, then |→b||b→| is equal to : [JEE Main 2019]
  10. Let →a=^i−2^j+^ka→=i^−2j^+k^ and →b=^i−^j+^kb→=i^−j^+k^ be two vectors. If →cc→ is a   [JEE Main 2020]
  11. If →aa→ and →bb→ are non-collinear vectors, then the value of a for which the vectors : [JEE Main 2013]
  12. If a unit vector →aa→ makes angles π/3π/3 with ^i,π/4i^,π/4 with ^jj^ and θ∈(0,π)θ∈(0,π) with ^kk^, then a value of θθ is : – [JEE Main 2019]
  13. The coordinates of the two points lying on….[WBJEE 2011]
  14. The distance of the point (1, 2, -4) from the….[KCET 2020]
  15. The midpoint of PR is _______…...[KCET 2010]

Sample Questions

Ques. Find the distance of the plane whose equation is given by 3x - 4y +12z =3, from the origin. (2 marks)

Ans. We can observe that the point here is the origin (0,0,0) while A= 3, B = -4, C= 12 and D = 3

By using the cartesian form; d = |Ax1+By1+Cz1−D / √A2+B2+C2|.

d = | (3 x 0) + (- 4 x 0) + (12 x 0) – 3 | / (32 + (-4)2 + (12)2)1/2

= 3 / (169)1/2

= 3 / 13 units is the required distance.

Ques. Find the distance of the plane whose equation is given by 4x - 3y +12z =3, from the origin. (2 marks)

Ans. we can observe that the point here is the origin (0,0,0) while A= 4, B = -3, C= 12 and D = 3

BY using the cartesian form; d = |Ax1+By1+Cz1−D / √A2+B2+C2|.

d = | (4 x 0) + (- 3 x 0) + (12 x 0) – 3 | / (42 + (-3)2 + (12)2)1/2

= 3 / (169)1/2

= 3 / 13 units is the required distance.

Ques. Drive the formula for calculating perpendicular distance from a point to a plane in vector form. (5 marks)

Ans. let us consider a point B to whom the position vector is given by \(\vec{r}\) and a plane P, which is given by the following equation,

 \(\vec{r}\). \(​​\vec{N}\) =d

where N is normal to the plane.

Now, consider another plane which is parallel to the first plane and that passes through the point B, by this we will get the equation of the second plane where N is normal to the plane. This equation can be written as,

(\(\vec{r}\) − \(\vec{a}\)). \(​​\vec{N}\) =0 

or in other easier way,

\(\vec{r}\). \(​​\vec{N}\) = \(\vec{a}\). \(​​\vec{N}\)

Now let’s consider O as the origin of the coordinates, the distance between the first plane from the origin is written as ON. Similarly, the distance of the other plane from the origin O is denoted as ON’. The distance between these two planes is calculated by ON–ON’. This distance is equal to

ON−ON′= d′ =|d−a. N^ |

We can also find the perpendicular distance of point B on the plane P’ from P. Therefore, we can conclude that for a plane which is denoted by the following equation \(\vec{r}\). \(​​\vec{N}\) =\(​​\vec{D}\) and a point B whose position vector we know, we can calculate the perpendicular distance from a point to the plane with the given formula 

d=| \(\vec{r}\). \(​​\vec{N}\) − \(\vec{D}\) / |\(​​\vec{N}\)|

Now if we want to calculate the length of the plane from the origin O, we need to substitute the value 0 in the place of the position vector. this will give us, 

d=|\(\vec{D}\) / \(​​\vec{N}\)|

Ques. Find the distance between the given plane 2x + 4y – 4z – 6 = 0 and the point M (0, 3, 6). (3 marks)

Ans. We can observe that the point here is (0,3,6) while A= 2, B = 4, C= -4 and D = 6

By using the cartesian form; d = |Ax1+By1+Cz1−D / √A2+B2+C2|.

d = | (2 x 0) + (4 x 3) + (-4x 6) – 6 | / (22 + (4)2 + (-4)2)1/2

= -18 / (36)1/2

= 18 / 6 units is the required distance.

or, 3 is the required distance.

Ques. Drive the formula for calculating perpendicular distance from a point to a plane in cartesian form. (5 marks)

Ans.  Suppose a plane that is denoted by the Cartesian equation,

Px + Qy + Rz = S

Now suppose that a point whose position vector and Cartesian coordinates of position vector are given by,

A (x1, y1, z1).

We can also write the position vector in the form

\(\vec{a}\) = x1 iˆ+ y1 jˆ +z1

For finding the distance of point B from the plane by using the vector form of the formula, we can find the normal vector to the plane, which is given as,

N =Aiˆ+ Bjˆ+ Ckˆ

When we use the formula, the perpendicular distance of point B from the given plane is written as,

d= |\(\vec{a}\). \(\vec{N}\) −\(\vec{D}\)/ |\(\vec{N}\) |

d = |Ax1+By1+Cz1−D / √A2+B2+C2|

This is the equation that will give us the perpendicular distance of a given point from its plane by using the Cartesian method.

Ques. What is the equation of a plane perpendicular to a vector? (2 marks)

Ans. The equation of a plane which is perpendicular to a vector and is passing through a point is written as A (x – x0) + B (y – y0) + C(z – z0) = 0, here (A, B, C) is the normal vector to the plane (x0, y0, z0).

Ques. What is the perpendicular distance of a point 3/4 passing through y axis? (2 marks)

Ans. To answer this particular question draw a graph which and jot down the points 3 by 4 on the graph then we will find out that the distance of the point will y axis coordinate with x axis. Hence, point 3 will be on the y axis whereas 4 will be on the x axis. hence the answer will be 3.

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