# 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 = x1  + y1 + z1
• Write direction vector using d.r.s a, b, c as  b = a   + b  + c
• 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 2  –  + 4 and direction of 2  –  + 4 . Also, reduce it to cartesian form.
• Solution:

The line passes through point with position vector 2  –  + 4

Let a = 2  –  + 4
The direction of the line is along 2  –  + 4
Let b =  2  –  + 4
The vector equation of the line is

r = a + λ b

r = (2  –  + 4 ) + λ(2  –  + 4 )

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 3  –  + 2. Also reduce it to cartesian form.
• Solution:

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

Its position vector is  a = 4  – 2 + 5
The direction of the line is along  3  –  + 2

Let b =  3  –  + 2
The vector equation of the line is

r = a + λ b

r = (4  – 2 + 5  ) + λ(3  –  + 2 )

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 4  –  + 2 and direction of – 2  +  + . Also, reduce it to cartesian form.
• Solution:

The line passes through point with position vector 4  –  + 2

Let a = 4  –  + 2
The direction of the line is along – 2  +  +

Let b =  – 2  +  +
The vector equation of the line is

r = a + λ b

r = (4  –  + 2) + λ(- 2  +  + )

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 = –   –  + 2
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 =  3  + 2 +
The vector equation of the line is

r = a + λ b

r = (-   –  + 2) + λ(3  + 2 + )

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 2  –  – 4 and direction of   – 2 + . Also, reduce it to cartesian form.
• Solution:

The line passes through point with position vector 2  –  – 4

Let a = 2  –  – 4
The direction of the line is along    – 2 +

Let b =   – 2 +

The vector equation of the line is

r = a + λ b

r = (2  –  – 4) + λ(   – 2 + )

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 2  – 3  + 4 and direction of 3   +4 – 5. Also, reduce it to cartesian form.
• Solution:

The line passes through point with position vector 2  – 3  + 4

Let a = 2  – 3  + 4
The direction of the line is along 3   + 4 – 5

Let b =3   + 4 – 5

The vector equation of the line is

r = a + λ b

r = (2  – 3  + 4) + λ(3   + 4 – 5)

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 = 5  + 4  + 3
The direction ratios are -3, 4, 2
Hence b = – 3  + 4  + 2

The vector equation of the line is

r = a + λ b

r = (5  + 4  + 3) + λ(- 3  + 4  + 2)

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 2  + 2  – 3. Also, reduce it to cartesian form.
• Solution:

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

Its position vector is a = 3  + 2  +

The direction of the line is along 2  + 2  – 3

Let b =  2  + 2  – 3

The vector equation of the line is

r = a + λ b

r = (3  + 2  + ) + λ( 2  + 2  – 3)

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 2  –   + 3. Also, reduce it to cartesian form.
• Solution:

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

Its position vector is a = 5  – 2  + 4

The direction of the line is along 2  –   + 3

Let b =  2  –   + 3

The vector equation of the line is

r = a + λ b

r = (5  – 2  + 4) + λ( 2  –   + 3)

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   –  2 + 3. Also, reduce it to cartesian form.
• Solution:

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

Its position vector is a =   + 2  + 3

The direction of the line is along    –  2  + 3

Let b =    –  2  + 3

The vector equation of the line is

r = a + λ b

r = (   + 2  + 3) + λ(   –  2  + 3)

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   +   – 2

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 =    +   +
The vector equation of the line is

r = a + λ b

r = ( -5   +   – 2) + λ(  +   + )

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

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   +   –  and is parallel to the line joining the points –   +   + 4  and    +  2 + 2. Also, reduce it to cartesian form.
• Solution:

The line passes through point with position vector is 2   +   –

Let  a = 2   +   –

p = 2   +   –

q = 2   +   –
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

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