#### Question

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.

#### Solution

Given equations of the line are:

3x -1 = 6y +2 = 1 - z

Rewriting the above equation, we have,

`3(x-1/3)=6(y+2/6)=-(z-1)`

`((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

This shows that the given line passes through (1/3, -1/3, 1)

Therefore, the given line passes through the point having

position vector `bara=1/3hati-i/3hatj+hatk` and is parallel to the

vector `barb=1/3hati+1/6hatj-hatk`

So its vector equation is

`barr=(1/3hati-1/3hatj+hatk)+lambda(1/3hati+1/6hatj-hatk)`