The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.
Given equations of the line are:
3x -1 = 6y +2 = 1 - z
Rewriting the above equation, we have,
Now consider the general equation of the line:
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
So its vector equation is