The cartesian equation of a line is ` 3x + 1 = 6 y - 2 = 1 -z`.
Find the vector equation of the line.

Explanation

The cartesian form of the equation of the line is
` 3 x + 1 = 6 y - 2 = 1 - z `
i.e., ` ( x + ( 1 ) /(3))/(((1) /(3))) = ( y - ( 1 ) /(3))/((( 1 ) /(6))) = ( z - 1 ) /(- 1 ) `
This line is passing through the point ` A ( - ( 1 ) /(3), ( 1 ) /(3), 1 ) ` and having direction ratios ` ( 1 ) /(3), ( 1) /(6) , - 1 , i.e., 2 , 1 - 6`.
Let ` bar a ` be the position vector of the point A w.r.t. the origin and ` bar b ` be the vector parallel to the line.
` therefore bar a = - ( 1 ) /(3) hati + ( 1) /(3) hatj + hatk and bar b = 2 hati + hatj - 6 hatk `
The vector equation of the line passing through ` A ( bar a ) ` and parallel to ` bar b ` is ` bar r = bar a + lamda bar b ` where ` lamda ` is a scalar.
` therefore ` the vector equation of the line is
` vec r = (- ( 1 ) /(3) hati + ( 1 ) /(3) hatj + hatk ) + lamda ( 2 hati + hatj - 6 hatk ) `