Distance between a Point and a Plane

In this post, i will be writing about how to find the distance of a point from a plane from the coordinates of the point and the equation of the plane. 

When we talk about the distance of a point to a plane we are talking about the shortest path connecting the point to the plane. The shortest path from a point to a plane is the path which is perpendicular to the plane. The length of the shortest path is what we are calling ‘the distance between the point and the plane’. If the point is ON the plane, then obviously the distance is zero.

In the above picture, QR is the shortest path from point Q to plane P, where point R on the plane is the foot of the perpendicular. 

Equation of P is 2x + y - 2z = 5. Let coordinates of the foot of the perpendicular be (a, b, c). Clearly, Q is not on P as the coordinates of Q does not satisfy the equation of P. So QR is a non-zero distance.

(6 - 1 - 4 = 1 ≠ 0).

Define a vector from Q to R(\(\overrightarrow{QR}\)) or from R to Q(\(\overrightarrow{RQ}\)), whichever way will do. 

I will define \(\overrightarrow{QR}\). It is a vector normal(perpendicular) to the plane. The magnitude of this vector is equal to the length of the segment QR, which is what we want to find. 

To write \(\overrightarrow{QR}\) in component form, we simply subtract the coordinates of its tail from the coordinates of its head. 

\(\overrightarrow{QR} = (a - 3)\hat{i} + (b + 1)\hat{j} + (c - 2)\hat{k}\)

We also have another normal vector to the plane which is embedded in the equation of the plane. The coefficients of x, y and z terms in the equation are the \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) components of that normal vector. So that normal vector from the equation is \(2\hat{i} + \hat{j} - 2\hat{k}\). I will use \(\overrightarrow{n}\) to denote it. 

Both the vectors \(\overrightarrow{n}\) and \(\overrightarrow{QR}\) are perpendicular to the plane, so they must be parallel to each other. 

Parallel vectors are scalar multiples of one another. This means the components of one vector can be written as some number(scalar) ‘t’ times the corresponding components of the other vector. 

a - 3 = t \(\times\) 2 = 2t

b + 1 = t \(\times\) 1 = t

c - 2 = t \(\times\) (-2) = -2t

We can write \(\overrightarrow{QR}\) as,

\(\overrightarrow{QR} = 2t \hat{i} + t \hat{j} - 2t \hat{k}\)

The value of ‘t’ can be easily determined from the equation of the plane. Note that point R(a, b, c) is on the plane, so it must satisfy the equation of the plane. 

2a + b - 2c = 5 

In place of ‘a’, ‘b’ and ‘c’, we can substitute ‘2t + 3’, ‘t - 1’ and ‘-2t + 2’. 

2(2t + 3) + t - 1 - 2(-2t +2) = 5

4t + 6 + t - 1 + 4t - 4 = 5

9t + 1 = 5

t = \(\frac{4}{9}\)

Now that we have the value of ‘t’, we can get the components of \(\overrightarrow{QR}\). 

\(\overrightarrow{QR} = \frac{8}{9}\hat{i} + \frac{4}{9}\hat{j} - \frac{8}{9}\hat{k}\)

The distance between the point and the plane is the magnitude of \(\overrightarrow{QR}\). 

\(|\overrightarrow{QR}| = \sqrt{(\frac{8}{9})^2 + (\frac{4}{9})^2 + (\frac{-8}{9})^2\)

    \( = \frac{1}{9} \sqrt{64 + 16 + 64}\)

   \( = \frac{12}{9} = \frac{4}{3}\) units

So the (shortest) distance from Q to the plane is about 1.33 units.