Class 12 Maths Three Dimensional Geometry | Equation of a line in Space |

**Equation of a line in Space**

A line is uniquely determined if:

- It passes through a given point and has direction, or
- It passes through two given points.

__Line through a given point and parallel to a given vector:-__

__Derivation of Cartesian form from vector form__

Let the coordinates of the given point A be (x_{1}, y_{1}, z_{1}) and the direction ratios of the line be a, b, c.

Let the coordinates of the any point P be (x, y, z).

Consider r^{->} = xî +yĵ +zk̂**,** a=x_{1}î+y_{1}ĵ+z_{1}k̂and b^{->}= (a î +bĵ +ck̂) line vector L^{->}=a î +bĵ +ck̂.

Equation of line: - r^{->}**= **P^{->}** + **λL^{->}**.**(equation(1))

Using equation (1) x_{1}î+y_{1}ĵ+z_{1}k̂= x_{1}î+y_{1}ĵ+z_{1}k̂+λ (aî +bĵ +ck̂).

Therefore, x=x_{1}+λa => (x-x_{1})/(a)=λ

y=y_{1}+λb => (y-y_{1})/(b) =λ

z=z_{1}+λc => (z-z_{1})/(c)=λ

Therefore Equation of line in Cartesian form is given by :-

**(x-x _{1})/ (a) = (y-y_{1})/ (b) = (z-z_{1})/(c)**(By eliminatingλ)

**Note: -** If l, m and n are direction cosines of the line then the equation of line is given as:

**=(x–x _{1})/l =(y-y_{1})/m= (z-z_{1})/n**

__Problem:__**-**

Find the equation of the line which passes through the point (1, 2, and 3) and is parallel tothe vector 3î+2ĵ -2k̂.

__Answer:-__

It is given that the line passes through the point A (1, 2, and 3). Therefore, the position vector through A is a^{->}=î+2ĵ +3k̂ and b^{->} =3î +2ĵ -2k̂.

It is known that the line which passes through point A and parallel to b^{->} is given by r^{->}**=**a^{->}**+ **λb^{->}

=>r^{->}** =**î +2ĵ +3k̂+λ (3î +2ĵ -2k̂).

This is the required equation of the line.

__Problem:-__

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î-ĵ +4k̂ and is in the direction î +2ĵ -k̂?

__Answer:-__

It is given that the line passes through the point with position vector a^{->}**=**2î -ĵ +4k̂(equation (1))

b^{->}** = **î +2ĵ -k̂(equation (2))

It is known that a line through a point with position vector a⃗and parallel tob ⃗ is givenby the equation, r^{->}** = **a^{->}** + **λb ^{->}**.**

=>r^{->}** =**2î -ĵ +4k̂**+ **λ(î +2ĵ -k̂)

This is the required equation of the line in vector form.

r^{->}** = **x î +yĵ +zk̂**=>**(λ+2)î**+** (2λ-1)ĵ**+** (-λ+4)k̂

Eliminating λ, we obtain the Cartesian form equation as:

(x-2)/ (1) =(y+2)/ (2) = (z-4)/ (-1)

This is the required equation of the given line in Cartesian form.

__Problem:-__

Find the Cartesian equation of the line which passes through the point(−2, 4, −5) and parallel to the line given by (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)?

__Answer:-__

It is given that the line passes through the point (−2, 4, −5) and is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)

The direction ratios of the line,(x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6), are 3, 5, and 6.

The required line is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6).

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x_{1}, y_{1}, z_{1}) and with direction ratios, a, b, c, is given by (x-x_{1})/ (a) =(y-y_{1})/ (b) = (z-z_{1})/(c)

Therefore the equation of the required line is (x+2)/ (3k) =(y-4)/ (5k) = (z+5)/ (6k)

=>(x+3)/ (3) = k

=> (y-4)/ (5) = k

=> (z+8)/ (6) =k.

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