Equation of Line in Space

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Science > Mathematics > Straight Line in SpaceYou are Here

Theorem – 1 (Vector Equation):

  • 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 Unit Vector i  + y Unit Vector j + z  Unit Vector k
  • Unit Vector kPosition 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 (ra) × b.

Theorem – 2 (Cartesian Equation):

  • 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

Type – I: To find Direction Ratios and 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



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



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

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.

Type – II A: Conversion of Vector Equation into Cartesian Equation

ALGORITHM

  1. Equate the R.H.S. vector form r = a + λ b to r = x Unit Vector i  + y Unit Vector j + z  Unit Vector k
  2. Open the brackets of R.H.S. group the terms
  3. Equate corresponding terms on both the sides
  4. Find three distinct equations for λ.
  5. Equate the three equations

Example – 05:

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

The vector equation of the line is   r = (2 Unit Vector i  –  Unit Vector j +  4Unit Vector k) + λ( Unit Vector i  +  Unit Vector j –  2Unit Vector k)

Where, r = x Unit Vector i  + y Unit Vector j + z  Unit Vector k

∴  x Unit Vector i  + y Unit Vector j + z  Unit Vector k = (2 Unit Vector i  –  Unit Vector j +  4Unit Vector k) + λ( Unit Vector i  +  Unit Vector j –  2Unit Vector k)

∴  x Unit Vector i  + y Unit Vector j + z  Unit Vector k = 2 Unit Vector i  –  Unit Vector j +  4Unit Vector k + λUnit Vector i  + λUnit Vector j –  2λUnit Vector k

∴  x Unit Vector i  + y Unit Vector j + z  Unit Vector k = (2 + λ) Unit Vector i  + (-1 + λ) Unit Vector j +  (4 – 2 λ)Unit Vector k



∴ 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

Type – II B: Conversion of Cartesian Equation into Vector Equation

ALGORITHM (Method – I)

  1. Write given the cartesian equation in standard form. 
  2. Then write the position vector of point through which line is passing. a = x1Unit Vector i  + y1Unit Vector j + z1Unit Vector k
  3. Direction ratios of the line are a, b, and c. Write the direction vector,  b = a Unit Vector i  + b Unit Vector j + c Unit Vector k
  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 = (x1Unit Vector i  + y1Unit Vector j + z1Unit Vector k)+ λ (a Unit Vector i  + b Unit Vector j + c Unit Vector k)

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 Unit Vector i  + y Unit Vector j + z  Unit Vector k
  4. Group terms on R.H.S without λ and with λ.
  5. Get the vector equation in the format r = (x1Unit Vector i  + y1Unit Vector j + z1Unit Vector k)+ λ (a Unit Vector i  + b Unit Vector j + c Unit Vector k)

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)Unit Vector i  – (1/3)Unit Vector j + 2Unit Vector k



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 = Unit Vector i  + 2Unit Vector j + 3Unit Vector k

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 Unit Vector i  + y Unit Vector j + z  Unit Vector k

 r = (3λ – 5) Unit Vector i  + (5λ – 4) Unit Vector j + (6λ – 5) Unit Vector k

 r = 3λ  Unit Vector i – 5 Unit Vector i   + 5λ Unit Vector j – 4 Unit Vector j + 6λ Unit Vector k – 5 Unit Vector k

 r =  – 5 Unit Vector i   – 4 Unit Vector j – 5 Unit Vector k + 3λ  Unit Vector i +  5λ Unit Vector j  + 6λ Unit Vector k

 r =  (- 5 Unit Vector i   – 4 Unit Vector j – 5 Unit Vector k ) + λ(3Unit Vector i +  5Unit Vector j  + 6Unit Vector k )

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 = 6Unit Vector i  – 4Unit Vector j + 5Unit Vector k

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

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

 r = a + λ b

 r = (6Unit Vector i  – 4Unit Vector j + 5Unit Vector k) + λ (2Unit Vector i  + 7Unit Vector j + 3Unit Vector k)

Where λ ∈ R, and is a scalar/parameter

Example – 9:

  • 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 = 5Unit Vector i  – 4Unit Vector j + 6Unit Vector k

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

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

 r = a + λ b

 r = (5Unit Vector i  – 4Unit Vector j + 6Unit Vector k) + λ (3Unit Vector i  + 7Unit Vector j + 2Unit Vector k)

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 = bUnit Vector i  + 0Unit Vector j + dUnit Vector k = bUnit Vector i  + dUnit Vector k

The d.r. s of the line are a, 1, c ≡ a, b, c
Hence, the direction vector of the line is  b = aUnit Vector i  + Unit Vector j + cUnit Vector k

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

r = a + λ b

 r = (bUnit Vector i  + dUnit Vector k) + λ (aUnit Vector i  + Unit Vector j + cUnit Vector k)

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)Unit Vector i  + (1/3)Unit Vector j + Unit Vector k

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)Unit Vector i  + (1/6)Unit Vector j – Unit Vector k

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 =  Unit Vector i  – (1/3)Unit Vector j + (1/3)Unit Vector k

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 Unit Vector i  + 2Unit Vector j + Unit Vector k

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

Cartesian equation

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

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 Unit Vector i  + Unit Vector j – 6Unit Vector k

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

Science > Mathematics > Straight Line in SpaceYou are Here
Physics Chemistry  Biology  Mathematics

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