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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 (x
_{1}, y_{1}, z_{1}) as a = x_{1}+ y_{1}+ z_{1} - 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(x_{1}, y_{1}, z_{1})

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

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

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

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

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

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

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

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

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

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

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

Science > Mathematics > Straight Line in Space > You are Here |

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