Equation of a Line in Space

Equation of a line through a given point and parallel to a given vector `vec b`:
Let `vec a` be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector `vec b` . 
Let  `vec r` be the position vector of an arbitrary point P on the line in following fig. 

Then `vec (AP)`  is parallel to the vector `vec b` , i.e., `vec (AP) `= `λ vec b` , where λ is some real number.
But  `vec (AP) = vec (OP) - vec (OA)`
i.e. `lambda vec b =  vec r - vec a`
Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line.  Hence, the vector equation of the line is given by 
`vec r = vec a + lambda vec b`    ...(1)

Remark:  If `vec b = a hat i + b hat j + c hat k ` , then a,b ,c are direction ratios of the line and conversely , if a, b , c are direction ratios of a line , then `vec b = a hat i +b hat j +c hat k`  will be the parallel to the line. Here, b should not be confused with `| vec b|`.

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. Consider the coordinates of any point P be (x, y, z). Then
`vec r = x hat i + y hat j + z hat k ;` 
`vec a = x_1 hat i + y_1 hat j + z_1 hat k`
and `vec b = a hat i + b hat j + c hat k`
Substituting these values in (1) and equating the coefficients of `hat i , hat j " and " hat k` , we get
`x = x_1 + lambda a` ; y = `y_1 +lambda  b` ; `z = z_1 + lambda c`  ...(2)
These are parametric equations of the line. Eliminating the parameter λ from (2), we get

`(x - x_1)/a = (y - y_1)/a = (z - z_1)/c`      ...(3)
This is the Cartesian equation of the line. 

Equation of a line passing through two given points:
Let `vec a` and `vec b` be the position vectors of two points `A(x_1 , y_1 , z_1)` and `B (x_2 , y_2 , z_2)` respectively that are lying on a line in following fig. 

Let `vec r`be the position vectors of an arbitrary point P(x , y , z), then P is a point on the  line if and only if `vec (AP) = vec r - vec a` and  ` vec (AB) = vec b - vec a` are collinear vectors. Therefore, P is on the line if and only if `vec r = vec a = lambda (vec b - vec a)`
or `vec r = vec a + lambda (vec b - vec a) , lambda ∈ R`   ...(1)
This is the vector equation of the line. 

Derivation of cartesian form from vector form  
We have 
`vec r = x hat i + y hat j + z hat k` , `vec a = x_1 hat i +y_1 hat j + z_1 hat k` and `vec b = x_2 hat i + y_2 hat j + z_2 hat k`,
Substituting these values in (1), we get 
`x hat i + y hat j+ z hat k = x_1hat i +y_1 hat j + z_1 hat k , + lambda [(x_2 - x_1) hat i +(y_2 - y_1) hat j + (z_2 -z_1 hat k)]`
Equating the like coefficients of `hat i , hat j , hat k ,`we have
`x = x_1 + lambda (x_2 - x_1); y = y_1 + lambda (y_2-y_1) ; z = z_1 +lambda (z_2 - z_1)`
On eliminating  λ, we obtain

`(x - x_1)/(x_2 - x_1) = (y - y_1)/ (y_2 - y_1) = (z - z_1) / (z_2 - z_1)`

which is the equation of the line in Cartesian form.
Video link : https://youtu.be/3GZQ8iiNvDU