Distance of a Point from a Plane

Distance between line and plane formula 

 The distance of a point from a plane can be calculated in vector form and cartesian Form. 

• Vector form – Let us consider a point B whose position vector is a and on plane 1. The equation of the plane is   r .   N  =  D . (Vector N is normal to the plane.)

 Now let’s consider another plane 2 through point B parallel to plane 1. The equation of the second plane 2 is (  r –   a   ). N =0. This can also be written as  r .N =a  .N.  

Therefore, the distance of plane two from its origin is a N . Thus, the distance from point B to plane 1 is d- a n→ .

• Cartesian Form – let us assume that point P (x1, y1, z1) has a position vector A. Let the cartesian equation of the given plane by Ax + By +Cz = D. 

Then, a = x1 i + y1 j+ z1 k

And N = A i + B j + C k 

In one of the earlier points, we have established that the distance of the point from the plane is the perpendicular drawn from the projection onto the plane. 

Therefore,( x1i + y1 j + z1 k) . (A i + B j + C k)/A2+B2+C2

= (Ax1+By1+Cz1-D)/A2+B2+C2

Distance of a point from plane solved examples. 

 Example 1- Find the distance between the  plane 6 i-3 j+2 k=4 and point (2, 5, -3). 

a=2 i+5 j-3k , N=6 i-3 j+2k, and d = 4. 

Since, (x1 i + y1 j + z1 k) . (A i + B j + C k-D)/A2+B2+C2

 = Ax1+By1+Cz1/A2+B2+C2

Therefore, (2 i+5 j-3k).(6 i-3 j+2k)=-9-4=-13

36+9+4 = 49

Therefore, the distance between point and plane is 13/7.  

 Example 2- Find the distance of a plane 3x – 4y + 12z = 3 from the point of its origin. 

A = 3, B = -4, C = 12, and D = 3

Since we are calculating the distance of a plane from the point of origin, the point is (0, 0, 0). 

Distance = d = (x1 i + y1 j + z1 k) . (A i + B j + C k)/A2+B2+C2 = Ax1+By1+Cz1/A2+B2+C2

 d = 3 ×0 +-4 ×0+12 ×0 -3=-3

A2+B2+C2=13

Therefore, the distance of point from plane = 3/13

 Example 3 – Find the distance between a point (3, -2, 1) and plane 2x – y + 2z + 3 = 0 

A = 2, B = -1, C = 2, and D = – 3

Distance = d = x1 i + y1 j + z1 k) × A i + B j + C k-D/A2+B2+C2 = Ax1+By1+Cz1/A2+B2+C2

d = 2 ×3 +-1 ×-2+2 ×1+ 322+ (-1)2+ 21 

d = 6 +2+2+ 34+1+4 = 139 

d = 133 

Therefore, the distance between a point (3, -2, 1) and plane 2x – y + 2z + 3 = 0 is 133 . 

 Example 4 – Find the distance between a point (2, 3, -5) and plane x + 2y – 2z = 9

A = 1, B = 2, C = -2, and D = 9

Distance = d = (x1 i + y1 j + z1 k) . (A i + B j + C k)/A2+B2+C2= Ax1+By1+Cz1/A2+B2+C2

 d = 1 ×2 +2 ×3 +-2 ×-5- 9=9

A2+B2+C2=3

Therefore, the distance between a point (2, 3, -5) and plane x + 2y – 2z = 9 is 3. 

 Example 5 – Find the distance between a point (-6, 0, 0) and plane 2x – 3y + 6z – 2 = 0

A = 2, B = -3, C = 6, and D = 2

Distance = d  = (x1 i + y1 j + z1 k) . (A i + B j + C k)/A2+B2+C2= Ax1+By1+Cz1/A2+B2+C2

d = 2 ×-6 +-3 ×0+6×0- 2=-14

A2+B2+C2=7 

Hence, distance =2 units

Other distance equations in three-dimension geometry 

• Distance between parallel lines 

The formula to determine the distance between two parallel lines is – 

d = PT = b ×(a2– a1)/|b|

• Distance between two skew lines 

Vector form – The shortest distance between two skew lines can be determined by 

d = |(b1 ×b2) .(a2– a1)|/|b1 × b2|

 Conclusion 

In this article we found different ways to calculate the distance between a plane and a line.We also took some examples to clear our concepts.Also we found how to calculate the distance between two parallel as well skew lines.