# 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.

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, (x_{1}, y_{1}, z_{1}) 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.

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.

*where (x, y, z) represents the coordinates of a variable point on the plane.*

**(**a**x +**b**y +**c**z +**d =**0****)****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:

πrepresents a plane in 3D space.vector is the position vector of a general point lying on the plane,

n̂is the unit vector normal to the plane anddis 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 anddis 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) / √(2

^{2}+ 1^{2}+ 2^{2})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) / √(4

^{2}+ 0^{2}+ 3^{2})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) / √(1

^{2}+ 3^{2}+ 4^{2})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:

πagain represents a plane in 3D space

l, m, nare the DC’s i.e. direction cosines of the the normal to the plane always satisfies this condition(l^{2}+ m^{2}+ n^{2}= 1)and

pis 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 whena, b, care 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 coises 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**√(a**.^{2 }+ b^{2}+ c^{2})

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 √(2

^{2}+ 1^{2}+ 2^{2}) = √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 √(1

^{2}+ 1^{2}+ 1^{2}) = √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 √(0

^{2}+ 1^{2}+ 3^{2}) = √(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

*in the cartesian form is defined as the length*

**π:**(a**x**+ b**y**+ c**z**+d = 0)*(*of the perpendicular drawn from that point to the plane.

**L**)

L = |a x_{o}+ b y_{o}+ c z_{o}+ d| / √(a^{2}+ b^{2}+ c^{2})

**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}= 0a = 2, b = 1, c = 2, d = 5

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

^{2}, 1^{2}, 2^{2})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}= 0a = 0, b = 3, c = 4, d = -7

L = |0 + (3 × 1) + (4 × 0) – 7| / √(0

^{2}, 3^{2}, 4^{2})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}= 1a = 4, b = 0, c = 3, d = 9

L = |(4 × 1) + (0 × 1) + (3 × 1) + 9| / √(4

^{2}, 0^{2}, 3^{2})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

|| = √(2

^{2}, 1^{2}, 2^{2}) = √9 = 3d = 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

|| = √(4

^{2}, 3^{2}, 0^{2}) = √(25) = 5d = 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

|| = √(1

^{2}, 3^{2}, 2^{2}) = √(14)d = 19 (given)

= (3 × 1) + (3 × 3) + (1 × 2) = 14

L = |14 – 19| / √(14)

L = 5 / √(14)