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

Explanation

Given equations of the line are :
` 3x - 1 = 6y + 2 = 1 - z`
` 3 (x - 1/3) = 6 (y + 2/6) = - (z - 1)`
` rArr" " ((x - 1/3))/(1/3)=((y + 1/3))/(1/6) = ((z-1))/(-1) ` ....(1)
Now consider the general equation of the line :
` (x -a)/l = (y -b)/m = (z -c)/n` ....(2)
Where l, m and n are the direction ratios of the line and the point (a, b, c) lies on the line.
Compare the equation (1), with the general equation (2)
We have, `l = 1/3 , m = 1/6 and n = - 1`
Also, ` a = 1/3 , b = - 1/3 and c = 1` ltbr. This shows that the given line passes through
`(1/3,1/3,1)`
Hence, `veca = 1/3 hat i - 1/3 hatj + hatk and vecb = 1/3 hati + 1/6 hat j - hat k`
The vector equation is given by :
` r = a + lambda b`
`r = (1/3 hati - 1/3 hatj + hatk) + lambda (1/3 hati+1/6 hatj-hatk)`