Equation of Line in Space

In this article, we shall study to write vector and cartesian equation of a line in Space.

Theorem – 1 (Vector Equation of Line in Space):

The vector equation of a straight line passing through a fixed point with position vector a  and parallel to a given vector b is r = a + λ b. Where λ is scalar and called the parameter.

Notes:

• In the above equation r is a position vector of any point P(x, y, z) on the line, then r = x   + y  + z  
• The position vector of any point on the line is taken as a + λ b. This form of the equation is called the vector form.
• If the line passes through the origin its vector equation is r = λ b
• The vector equation of a line passing through a fixed point with position vector a and parallel to a given vector b is also given as

(r – a) × b

Theorem – 2 (Cartesian Equation of Line in Space):

The cartesian equation of a straight line passing through a fixed point P(x₁, y₁, z₁) and having direction ratios (d.r.s) proportional to a, b, c respectively is given by 

Notes:

• If then x = aλ + x₁, y = bλ + y₁, and z = cλ + z₁. These equations are called the parametric equations of the line.
• The coordinates of any point on the line are (aλ + x₁, bλ + y₁, cλ + z₁).
• If the line passes through the origin its equation is

• Since the direction cosines (d.c.s) of a line are direction ratios (d.r.s) of the line, the equation of the line passing through a fixed point P(x₁, y₁, z₁) and having direction cosines (d.c.s) l, m, n respectively is given by

• The x-axis, y-axis, and z-axis pass through origin.
• d.r.s of x-axis are 1, 0, 0. Hence its equation is y = 0 and z = 0.
• d.r.s ofy-axiss are 0, 1, 0. Hence its equation is x = 0 and z = 0.
• d.r.s of z – axis are 0, 0, 1. Hence its equation is x = 0 and y = 0.

Theorem – 3:

The vector equation of a straight line passing through two fixed points with position vector a and b  is

r = a + λ( b – a)

Where λ is scalar and called the parameter.

Theorem – 4:

The cartesian equation of a straight line passing through two fixed points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by

To find Direction Ratios and Direction Cosines of Line in Space:

Algorithm:

• Write the equation in standard form
• Make sure there are no coefficients for the x, y, and z terms. If coefficients are present divide the numerator and denominator by the coefficient.
• Do simplification if any
• Then the denominators indicate the direction ratios.
• Using direction ratios, find direction cosines

Example 01:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 3/2, 0 i.e. 4, 3, 0 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are 4/5, 3/5, 0

Example 02:

• Find the direction cosines of the line 
• Solution:

The equation of the line is 

Writing in a standard form

The d.r.s of lines are -2, 6, -3 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are -2/7, 6/7, -3/7

Example 03:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 3, 0 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are 2/√13, 3/√13, 0.

Example 04:

• Find the direction cosines of the line 
• Solution:

The equation of the line is

Writing in a standard form

The d.r.s of lines are 2, 4, 3 ≡ a, b, c
Now d.c. s of the line are

Ans: d.c.s of the line are -2/√17, 2/√17, -3/√17.

Conversion of Vector Equation into Cartesian Equation of Line in Space:

Algorithm:

1. Equate the vector form r = a + λ b to r = x   + y  + z 
2. Open the brackets of R.H.S.
3. Group the terms of , , and 
4. Equate corresponding terms on both the sides
5. Find three distinct equations for λ.
6. Equate the three equations

Example 05:

• Find the cartesian equations of a line whose vector equation is  r = (2   –   +  4) + λ(   +   –  2)
• Solution:

The vector equation of the line is   

r = (2   –   +  4) + λ(   +   –  2)

Where, r = x   + y  + z 

∴  x   + y  + z  = (2   –   +  4) + λ(   +   –  2)

∴  x   + y  + z   = 2   –   +  4 + λ  + λ –  2λ

∴  x   + y  + z   = (2 + λ)   + (-1 + λ)  +  (4 – 2 λ)

∴ x = 2 + λ  ⇒  λ = x – 2 …………. (1)

∴ y = -1 + λ  ⇒  λ = y + 1 …………. (2)

∴ z = 4 – 2λ  ⇒  λ = (z – 4)/(-2)  …………. (3)

From equations (1), (2) and (3)

x – 2 = y + 1 = (z – 4)/(-2)

Thus the cartesian equations of lines are

Conversion of Cartesian Equation into Vector Equation of Line in Space:

Algorithm (Method – I):

1. Write given the cartesian equation in standard form. 
2. Then write the position vector of the point through which the line is passing. a = x₁  + y₁ + z₁
3. The direction ratios of the line are a, b, and c. Write the direction vector,  b = a   + b  + c 
4. Write the vector form of the equation as r = a + λ b. Where λ ∈ R, and is a scalar/parameter
5. Thus vector equation of line is r = (x₁ + y₁ + z₁)+ λ (a  + b  + c )

Algorithm (Method – II):

1. Let 
2. Find x = aλ + x₁, y = bλ + y₁, and z = cλ + z₁,
3. Substitute values of a, y, and z in the equation r = x   + y  + z  
4. Group terms on R.H.S without λ and with λ.
5. Get the vector equation in the format r = (x₁ + y₁ + z₁)+ λ (a  + b  + c )

Example 06:

• Find the vector equation of a line whose cartesian equation is  6x – 2 = 3y + 1 = 2z – 2
• Solution (Method – I):

The cartesian equation of the line is

Thus the line passes through the point (1/3, -1/3, 2)
The position vector of this point is  a =(1/3)  – (1/3) + 2

The d.r. s of the line are i.e. 1, 2, 3 ≡ a, b, c
Hence, the direction vector of the line is b =   + 2 + 3

Now, the vector form of the equation of the line is given by

 r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 07:

• Find the vector equation of a line whose cartesian equation is 
• Solution (Method – II):

The equation of the line is 

Let,   = λ
Thus x = 3λ – 5, y = 5λ – 4 and z = 6λ – 5
Now, r = x   + y  + z 

 r = (3λ – 5)   + (5λ – 4)  + (6λ – 5) 

 r = 3λ   – 5    + 5λ  – 4  + 6λ  – 5 

 r =  – 5    – 4  – 5  + 3λ   +  5λ   + 6λ 

 r =  (- 5    – 4  – 5  ) + λ(3 +  5  + 6  )

Where λ ∈ R, and is a scalar/parameter

Example 08:

• Find the vector equation of a line whose cartesian equation is 
• Solution:

The equation of the line is 

Thus the line passes through the point (6, -4, 5)
The position vector of this point is  a = 6  – 4 + 5

The d.r. s of the line are 2, 7, 3≡ a, b, c
Hence, the direction vector of the line is  b = 2  + 7 + 3

Now, the vector form of the equation of the line is given by

 r = a + λ b

 r = (6  – 4 + 5) + λ (2  + 7 + 3)

Where λ ∈ R, and is a scalar/parameter

Example 09:

• Find the vector equation of a line whose cartesian equation is 
• Solution:

The cartesian equation of the line is 

Thus the line passes through the point (5, -4, 6)
The position vector of this point is  a = 5  – 4 + 6

The d.r. s of the line are 3, 7, 2 ≡ a, b, c
Hence, the direction vector of the line is b = 3  + 7 + 2

Now, the vector form of the equation of the line is given by

 r = a + λ b

 r = (5  – 4 + 6) + λ (3  + 7 + 2)

Where λ ∈ R, and is a scalar/parameter

Example 10:

• Find the vector equation of a line whose cartesian equation is x = ay + b and z = cy + d . Find its direction ratios.
• Solution:

The cartesian equation of the line is

Thus the line passes through the point(b, 0, d)
The position vector of this point is  a = b  + 0 + d = b  + d

The d.r. s of the line are a, 1, c ≡ a, b, c
Hence, the direction vector of the line is  b = a  +  + c

Now, the vector form of the equation of the line is given by

r = a + λ b

 r = (b  + d) + λ (a  +  + c)

Where λ ∈ R, and is a scalar/parameter

Example 11:

• Find the vector equation of a line whose cartesian equation is 3x + 1 = 6y – 2 = 1- z . Find the fixed point through which it passes and its. d.r.s
• Solution:

The equation of the line is 3x + 1 = 6y – 2 = 1- z

Thus the line passes through the point ( -1/3, 1/3, 1)
The position vector of this point is a = (-1/3)  + (1/3) +

The d.r. s of the line are i.e. 1/3, 1/6, -1 ≡ a, b, c
Hence, the direction vector of the line is B = (1/3)  + (1/6) –

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 12:

• Find the vector equation of a line whose cartesian equation is 2x – 2 = 3y + 1 = 6z – 2. Find the fixed point through which it passes and its. d.r.s.
• Solution:

The equation of the line is 2x – 2 = 3y + 1 = 6z – 2

Thus the line passes through the point (1, – 1/3, 1/3)
The position vector of this point is a =    – (1/3) + (1/3)

The d.r. s of the line are i.e. 3, 2, 1≡ a, b, c
Hence, the direction vector of the line is b = 3   + 2 +

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter

Example 13:

• Find the vector equation of a line whose cartesian equation is 3x – 1 = 6y + 2 = 1 – z
• Solution:

The cartesian equation of the line is 3x – 1 = 6y + 2 = 1 – z

Thus the line passes through the point (1, – 1/3, 1)
The position vector of this point is a =    – (1/3) +

The d.r. s of the line are i.e. 1/3, 1/6, – 1 i.e. 2, 1, -6 ≡ a, b, c
Hence, the direction vector of the line is b = 2   +  – 6

Now, the vector form of the equation of the line is given by

r = a + λ b

Where λ ∈ R, and is a scalar/parameter