Physics |
Chemistry |
Biology |
Mathematics |

Science > Mathematics > Straight Line in Space > You 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 + y + z
- 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):

- The cartesian equation of a straight line passing through a fixed point P(x
_{1}, y_{1}, z_{1}) and having direction ratios (d.r.s) proportional to a, b, c respectively is given by

#### Notes:

- If then x = aλ + x
_{1}, y = bλ + y_{1}, and z = cλ + z_{1}. These equations are called the parametric equations of the line. - The coordinates of any point on the line are (aλ + x
_{1}, bλ + y_{1}, cλ + z_{1}). - 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(x
_{1}, y_{1}, z_{1}) 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(x
_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) 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

- Equate the R.H.S. vector form r = a + λ b to r = x + y + z
- Open the brackets of R.H.S. group the terms
- Equate corresponding terms on both the sides
- Find three distinct equations for λ.
- 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

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

#### ALGORITHM (Method – I)

- Write given the cartesian equation in standard form.
- Then write the position vector of point through which line is passing. a = x
_{1}+ y_{1}+ z_{1} - Direction ratios of the line are a, b, and c. Write the direction vector, b = a + b + c
- Write the vector form of the equation as r = a + λ b . Where λ ∈ R, and is a scalar/parameter
- Thus vector equation of line is r = (x
_{1}+ y_{1}+ z_{1})+ λ (a + b + c )

#### ALGORITHM (Method – II):

- Let
- Find x = aλ + x
_{1}, y = bλ + y_{1}, and z = cλ + z_{1}, - Substitute values of a, y, and z in the equation r = x + y + z
- Group terms on R.H.S without λ and with λ.
- Get the vector equation in the format r = (x
_{1}+ y_{1}+ z_{1})+ 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 – 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 = 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

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

Physics |
Chemistry |
Biology |
Mathematics |