Points, Lines and Planes

A Point in three-dimensional geometry is defined as a location in 3D space that is uniquely defined by an ordered triplet (x, y, z) where x, y, & z are the distances of the point from the X-axis, Y-axis, and Z-axis respectively.

Point in 3D space

A Line in three-dimensional geometry is defined as a set of points in 3D that extends infinitely in both directions and is represented by L : (x – x1) / l = (y – y1) / m = (z – z1) / n; here (x, y, z) are the position coordinates of any variable point lying on the line, (x₁, y₁, z₁) are the position coordinates of a point P lying on the line, and l, m, & n are the direction ratios (DRs). In 3D a line is also formed by the intersection of two non-parallel planes.

Line in 3D space

A Plane in three-dimensional (3D) geometry can be considered as a surface such that the line segment joining any two points on the surface lies completely on it. The general form of a plane in 3D is a first-degree equation in x, y, z i.e. (a x + b y + c z + d = 0) where (x, y, z) represents the coordinates of a variable point on the plane.

Plane in 3D space

Vector Form of The Equation of a Plane in Normal Form

The vector form of the equation of a plane in normal form is given by:

π:  . n̂  = d

Where:

π represents a plane in 3D space.

 vector is the position vector of a general point lying on the plane, n̂ is the unit vector normal to the plane and d is the distance of a plane from the origin.

Note: The vector equation of the plane in the form ( .  = d) is said to be in the normal form only when  is a unit vector normal to the plane and d is the distance of the plane from the origin. If  is not a unit vector then we have to divide the above equation by || on both the sides in order to convert it into the normal form.  or .n̂ = 

Examples

Example 1: The vector equation of the plane in 3D space which is at a distance of  8 units from the origin and normal to the vector (2 i+ j + 2 k) is given by?

Solution:

d = 8 and   = (2 i+ j + 2 k)

n̂ = (2 i+ j + 2 k) / √(2² + 1² + 2²) 

n̂ = (2 i+ j + 2 k) / √9

n̂ = (2/3) i+ (1/3) j + (2/3) k

Hence the required vector equation of the plane in normal form is

 . ((2/3) i+ (1/3) j + (2/3) k) = 8 which can be simplified as . (2 i+ 1 j + 2 k) = 24

Example 2: The vector equation of the plane in 3D space which is at a distance of  5 units from the origin and normal to the vector (4 i+ 3 k) is given by?

Solution:

d = 5 and   = (4 i+ 3 k)

n̂ = (4 i + 3 k) / √(4² + 0² + 3²)

n̂ = (4 i+ 3 k) / √(25)

n̂ = (4/5) i+ (3/5) k

Hence the required vector equation of the plane in normal form is  . ((4/5) i+ (3/5) k) = 5 which can be simplified as  . (4 i+ 3 k) = 25

Example 3: The vector equation of the plane in 3D space which is at a distance of  19 units from the origin and normal to the vector (i + 3 j + 4 k) is given by?

Solution:

d = 19 and   = (i+ 3 j + 4 k)

n̂ = (i+ 3 j + 4 k) / √(1² + 3² + 4²)

n̂ = (i+ 3 j + 4 k) / √(26)

Hence the required vector equation of the plane in normal form is  . (i+ 3 j + 4 k)/ (√(26)) = 19

Cartesian Form of The Equation of a Plane in Normal Form

The cartesian form of the equation of a plane in normal form is given by:

π: lx + my + nz = p

Where:

π again represents a plane in 3D space

l, m, n are the DC’s i.e. direction cosines of the normal to the plane always satisfies this condition (l² + m² + n² = 1)

and p is the distance of the plane from the origin

Note: Any cartesian equation of the plane in the form (ax + by + cz + d = 0) is said to be in the normal form only when a, b, c are the direction cosines of the normal to the plane and |d| is the distance of the plane from the origin.

If a, b, c are not the direction  cosines of the normal to the plane then we have to follow these steps:

• Step 1: Keep the terms of x, y, and z on the LHS and take the constant term d on the RHS.
• Step 2: If the constant term on the RHS is negative then make it positive by multiplying with (-1) on both sides of the equation.
• Step 3: Divide term on the both sides of the equation by √(a² + b² + c²).

After applying these steps the coefficients of x, y, and z on the LHS will become the direction cosines of the normal to the plane and the constant term on the RHS will become the distance of the plane from the origin.

Examples

Example 1: A plane in the 3D space is represented by (2x + y + 2z – 24 = 0) then the cartesian equation of this plane in the normal form is given by?

Solution:

Taking the constant term on the RHS

2x + y + 2z = 24

Dividing both sides of the above  equation by √(2² + 1² + 2²) = √9 = 3

(2/3) x + (1/3) y + (2/3) z = 8

Here l = 2 / 3 , m = 1 / 3 , n = 2 / 3 are the direction cosines & p = 8 is the distance from the origin.

Example 2: (x + y – z – 1 = 0) is a plane in 3D space then the cartesian equation of this plane in the normal form is given by?

Solution:

Taking the constant term on the RHS

x + y – z = 1

Dividing both sides of the above  equation by √(1² + 1² + 1²) = √3

(1/(√3)) x + (1/(√3)) y – (1/(√3)) z = 1 /√3

Here l = 1 / (√3) , m = 1 / (√3) , n = (-1) / (√3) are the direction cosines & p = 1 / √3 is the distance from the origin.

Example 3: A plane in the 3D space is given as (y + 3 z – 10 = 0) then the cartesian equation of this plane in the normal form is given by?

Solution:

Taking the constant term on the RHS

y + 3 z = 10

Dividing both sides of the above  equation by √(0² + 1² + 3²) = √(10)

(0/(√10)) x + (1/(√10)) y + (3/(√10)) z = 10 / √(10)

0 x + (1 / √(10)) y + (3 / (√10)) z = √(10)

Here l = 0 , m = 1 / √(10), n = 3 / √(10) are the direction cosines & p = √(10) is the distance from the origin.

Distance of a Point from a Plane in Cartesian Form

The distance of a point P (x_o, y_o, z_o) from a plane π: (a x + b y + c z +d = 0) in the cartesian form is defined as the length (L) of the perpendicular drawn from that point to the plane.

L = |a x_o + b y_o + c z_o + d| / √(a² + b² + c²)

Examples

Example 1: The distance of the point (2, 1, 0) from the plane (2 x + y + 2 z + 5 = 0) is given by?

Solution:

x_o = 2, y_o = 1, z_o = 0

a = 2, b = 1, c = 2, d = 5

L = |(2 × 2) + (1 × 1) + (0 × 2) + 5| / √(2², 1², 2²)

L = 10 / √9

L = 10 / 3

Example 2: The distance of the point (0, 1, 0) from the plane ( 3 y + 4 z = 7) is given by?

Solution:

x_o = 0, y_o = 1, z_o = 0

a = 0, b = 3, c = 4, d = -7

L = |0 + (3 × 1) + (4 × 0) – 7| / √(0², 3², 4²)

L = |3 – 7| / √(25)

L = 4 / 5

Example 3: The distance of the point (1, 1, 1) from the plane (4 x + 3 z + 9 = 0) is given by?

Solution:

x_o = 1, y_o = 1, z_o = 1

a = 4, b = 0, c = 3, d = 9

L = |(4 × 1) + (0 × 1) + (3 × 1) + 9| / √(4², 0², 3²)

L = |7 + 9| / √(25)

L = 16 / 5

Distance of a Point from a Plane in Vector Form

The distance of a point P having position vector from a plane π:  in vector form is defined as the length (L) of the perpendicular drawn from that point to the plane.

L = 

Examples

Example 1: The distance of a point with position vector (2 i + j + 0 k) from the plane . (2 i + j + 2 k) = 5 is given by?

Solution:

 = 2 i + j + 0 k

 = 2 i + j + 2 k

|| = √(2², 1², 2²) = √9 = 3 

d = 5 (given)

 = (2 × 2) + (1 × 1) + (0 × 2) = 5

L = |5 – 5| / 3

L = 0

Example 2: The distance of a point (5, 3, 0) from the plane . ( 4i + 3j) = 8 is given by?

Solution:

 = 5 i +3 j + 0 k

 = 4 i +3 j + 0 k

|| = √(4², 3², 0²) = √(25) = 5

d = 8 (given)

 .  = (5 × 4) + (3 × 3) + (0 × 0) = 29

L = |29 – 8| / 5

L = 21 / 5

Example 3: The distance of a point (3, 3, 1) from the plane . ( i + 3 j + 2 k) = 19 is given by?

Solution:

 = 3 i +3 j + k

 = i +3 j + 2 k

|| = √(1², 3², 2²) = √(14)

d = 19 (given)

= (3 × 1) + (3 × 3) + (1 × 2) = 14

L = |14 – 19| / √(14)

L = 5 / √(14)