#### notes

**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