Vector Equation of Line in Space

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Type – III A: To Find Vector Equation of Line Passing Through Given Point and Parallel to Given Line / Vector

ALGORITHM

  • Write position vector for a given point (x1, y1, z1) as a = x1Unit Vector i  + y1Unit Vector j + z1Unit Vector k
  • Write direction vector using d.r.s a, b, c as  b = a Unit Vector i  + b Unit Vector j + c Unit Vector k
  • The vector equation of the line is r = a + λ b. Where λ ∈ R, and is a scalar/parameter

Example – 14:

  • Find the vector equations of a line which passes through point with position vector 2Unit Vector i  – Unit Vector j + 4Unit Vector k and direction of 2Unit Vector i  – Unit Vector j + 4Unit Vector k . Also, reduce it to cartesian form.
  • Solution:

The line passes through point with position vector 2Unit Vector i  – Unit Vector j + 4Unit Vector k

Let a = 2Unit Vector i  – Unit Vector j + 4Unit Vector k
The direction of the line is along 2Unit Vector i  – Unit Vector j + 4Unit Vector k
Let b =  2Unit Vector i  – Unit Vector j + 4Unit Vector k
The vector equation of the line is

r = a + λ b

r = (2Unit Vector i  – Unit Vector j + 4Unit Vector k ) + λ(2Unit Vector i  – Unit Vector j + 4Unit Vector k )



Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(2, -1, 4) ≡ P(x1, y1, z1)
and its d.r.s are 1, 1, -2  ≡ a, b, c
The Cartesian form of the equation is

Example – 15:

  • Find the vector equations of a line which passes through a point (4, -2, 5) and parallel to the vector 3Unit Vector i  – Unit Vector j + 2Unit Vector k. Also reduce it to cartesian form.
  • Solution:

The line passes through point (4, -2 5)

Its position vector is  a = 4Unit Vector i  – 2Unit Vector j + 5Unit Vector k
The direction of the line is along  3Unit Vector i  – Unit Vector j + 2Unit Vector k



Let b =  3Unit Vector i  – Unit Vector j + 2Unit Vector k
The vector equation of the line is

r = a + λ b

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

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(4, -3, 5) ≡ P(x1, y1, z1)
and its d.r.s are 3, -1, 2  ≡ a, b, c
The Cartesian form of the equation is



Example 16:

  • Find the vector equations of a line which passes through point with position vector 4Unit Vector i  – Unit Vector j + 2Unit Vector k and direction of – 2Unit Vector i  + Unit Vector j + Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

The line passes through point with position vector 4Unit Vector i  – Unit Vector j + 2Unit Vector k

Let a = 4Unit Vector i  – Unit Vector j + 2Unit Vector k
The direction of the line is along – 2Unit Vector i  + Unit Vector j + Unit Vector k

Let b =  – 2Unit Vector i  + Unit Vector j + Unit Vector k
The vector equation of the line is

r = a + λ b

r = (4Unit Vector i  – Unit Vector j + 2Unit Vector k) + λ(- 2Unit Vector i  + Unit Vector j + Unit Vector k )



Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(4, -1, 2) ≡ P(x1, y1, z1)
and its d.r.s are -2, 1, 1  ≡ a, b, c
The Cartesian form of the equation is

Example – 17:

  • Find the vector equations of a line which passes through point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2. Also, reduce it to cartesian form.
    Solution:

The line passes through the point (-1, -1, 2).

Its position vector is a = – Unit Vector i  – Unit Vector j + 2Unit Vector k
The equation of given line is 2x – 2 = 3y + 1 = 6z – 2



Thus the direction ratios of the line are 1/2, 1/3 and 1/6 i.e 3, 2, 1 ≡ a, b, c
Now, the required line is parallel to the given line. Hence their d.r.s should be proportional.
Thus the direction of the required line is along

b =  3Unit Vector i  + 2Unit Vector j + Unit Vector k
The vector equation of the line is

r = a + λ b

r = (- Unit Vector i  – Unit Vector j + 2Unit Vector k) + λ(3Unit Vector i  + 2Unit Vector j + Unit Vector k)



Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(-1, -1, 2) ≡ P(x1, y1, z1)
and its d.r.s are 3, 2, 1  ≡ a, b, c
The Cartesian form of the equation is

Example – 18:

  • Find the vector equations of a line which passes through point with position vector 2Unit Vector i  – Unit Vector j – 4Unit Vector k and direction of Unit Vector i  – 2Unit Vector j + Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

The line passes through point with position vector 2Unit Vector i  – Unit Vector j – 4Unit Vector k

Let a = 2Unit Vector i  – Unit Vector j – 4Unit Vector k
The direction of the line is along  Unit Vector i  – 2Unit Vector j + Unit Vector k

Let b = Unit Vector i  – 2Unit Vector j + Unit Vector k

The vector equation of the line is



r = a + λ b

r = (2Unit Vector i  – Unit Vector j – 4Unit Vector k) + λ( Unit Vector i  – 2Unit Vector j + Unit Vector k)

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(2, -1, -4) ≡ P(x1, y1, z1)
and its d.r.s are 1, -2, 1  ≡ a, b, c
The Cartesian form of the equation is



Example – 19:

  • Find the vector equations of a line which passes through point with position vector 2Unit Vector i  – 3 Unit Vector j + 4Unit Vector k and direction of 3 Unit Vector i  +4Unit Vector j – 5Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

The line passes through point with position vector 2Unit Vector i  – 3 Unit Vector j + 4Unit Vector k

Let a = 2Unit Vector i  – 3 Unit Vector j + 4Unit Vector k
The direction of the line is along 3 Unit Vector i  + 4Unit Vector j – 5Unit Vector k

Let b =3 Unit Vector i  + 4Unit Vector j – 5Unit Vector k

The vector equation of the line is

r = a + λ b

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



Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(2, -3, 4) ≡ P(x1, y1, z1)
and its d.r.s are 3, 4, 5  ≡ a, b, c
The Cartesian form of the equation is

Example – 20:

  • Find the vector equations of a line which passes through point (5, 4, 3) and direction ratios -3, 4, 2 . Also reduce it to cartesian form.
  • Solution:

 The line passes through point (5, 4, 3).

Its position vector is a = 5Unit Vector i  + 4 Unit Vector j + 3Unit Vector k
The direction ratios are -3, 4, 2
Hence b = – 3Unit Vector i  + 4 Unit Vector j + 2Unit Vector k

The vector equation of the line is

r = a + λ b

r = (5Unit Vector i  + 4 Unit Vector j + 3Unit Vector k) + λ(- 3Unit Vector i  + 4 Unit Vector j + 2Unit Vector k)

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(5, 4, 3) ≡ P(x1, y1, z1)
and its d.r.s are -3, 4, 2  ≡ a, b, c
The Cartesian form of the equation is

Example – 21:

  • Find the vector equations of a line which passes through the point (3, 2, 1) and is parallel to the vector 2Unit Vector i  + 2 Unit Vector j – 3Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

 The line passes through the point (3, 2, 1).

Its position vector is a = 3Unit Vector i  + 2 Unit Vector j + Unit Vector k

The direction of the line is along 2Unit Vector i  + 2 Unit Vector j – 3Unit Vector k

Let b =  2Unit Vector i  + 2 Unit Vector j – 3Unit Vector k

The vector equation of the line is

r = a + λ b

r = (3Unit Vector i  + 2 Unit Vector j + Unit Vector k) + λ( 2Unit Vector i  + 2 Unit Vector j – 3Unit Vector k)

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(3, 2, 1) ≡ P(x1, y1, z1)
and its d.r.s are 2, 2, -3  ≡ a, b, c
The Cartesian form of the equation is

Example – 22:

  • Find the vector equations of a line which passes through the point (5, -2, 4) and is parallel to the vector 2Unit Vector i  –  Unit Vector j + 3Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

 The line passes through the point (5, -2, 4).

Its position vector is a = 5Unit Vector i  – 2 Unit Vector j + 4Unit Vector k

The direction of the line is along 2Unit Vector i  –  Unit Vector j + 3Unit Vector k

Let b =  2Unit Vector i  –  Unit Vector j + 3Unit Vector k

The vector equation of the line is

r = a + λ b

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

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(5, -2, 4) ≡ P(x1, y1, z1)
and its d.r.s are 2, -1, 3  ≡ a, b, c
The Cartesian form of the equation is

Example – 23:

  • Find the vector equation of a line which passes through the point (1, 2, 3) and is parallel to the vector Unit Vector i  –  2Unit Vector j + 3Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

The line passes through the point (1, 2, 3).

Its position vector is a = Unit Vector i  + 2 Unit Vector j + 3Unit Vector k

The direction of the line is along  Unit Vector i  –  2 Unit Vector j + 3Unit Vector k

Let b =  Unit Vector i  –  2 Unit Vector j + 3Unit Vector k

The vector equation of the line is

r = a + λ b

r = ( Unit Vector i  + 2 Unit Vector j + 3Unit Vector k) + λ( Unit Vector i  –  2 Unit Vector j + 3Unit Vector k)

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(1, 2, 3) ≡ P(x1, y1, z1)
and its d.r.s are 1, -2, 3  ≡ a, b, c
The Cartesian form of the equation is

Example – 24:

  • Find the vector equation of a line which is equally inclined to co-ordinate axes and passing through the point (-5, 1, -2). Also, reduce it to cartesian form.
  • Solution:

The line passes through the point (-5, 1, -2).

Its position vector is  a = -5 Unit Vector i  +  Unit Vector j – 2Unit Vector k

Now the line is equally inclined to co-ordinate axes,

hence its direction ratios of the line are 1, 1, 1  ≡ a, b, c
Let b =  Unit Vector i  +  Unit Vector j + Unit Vector k
The vector equation of the line is

r = a + λ b

r = ( -5 Unit Vector i  +  Unit Vector j – 2Unit Vector k) + λ(Unit Vector i  +  Unit Vector j + Unit Vector k)

Where λ ∈ R, and is a scalar/parameter
Now the line passes through the point
P(-5, 1, -2) ≡ P(x1, y1, z1)
and its d.r.s are 1, 1, 1  ≡ a, b, c
The Cartesian form of the equation is

Vector equation

i.e x + 5 = y – 1 = z +2

Example – 25:

  • Find the vector equation of a line which passes through point whose position vector is 2 Unit Vector i  +  Unit Vector j – Unit Vector k and is parallel to the line joining the points – Unit Vector i  +  Unit Vector j + 4Unit Vector k  and  Unit Vector i  +  2Unit Vector j + 2Unit Vector k. Also, reduce it to cartesian form.
  • Solution:

The line passes through point with position vector is 2 Unit Vector i  +  Unit Vector j – Unit Vector k

Let  a = 2 Unit Vector i  +  Unit Vector j – Unit Vector k

 p = 2 Unit Vector i  +  Unit Vector j – Unit Vector k

 q = 2 Unit Vector i  +  Unit Vector j – Unit Vector k
Now,

Now the line is parallel to
The direction ratios of line are 2, 1, -2
Let
The vector equation of the line is

Where and is scalar / parameter.
Now the line passes through the point
P(2, 1, -1) P(x1, y1, z1)
and its d.r.s are 2, 1,-2 a, b, c
The cartesian form of equation is

Example – 26 (H):
Find the vector equations of a line which passes through point whose position vector is and is parallel to the line joining the points and . Also reduce it to cartesian form. (CBSE 1995 C, 2003)
Solution:
The line passes through point with position vector
Let and
Now,

Now the line is parallel to
The direction ratios of line are 2, -2, 1
Let
The vector equation of the line is

Where and is scalar / parameter.
Now the line passes through the point
P(2, -1, 1) P(x1, y1, z1)
and its d.r.s are 2, -2, 1 a, b, c
The cartesian form of equation is

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

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