The Cartesian equationof a line are `3x+1=6y=2=1-z`. Find the direction ratios and write down its equation in vector form.

Explanation

The Cartesian equation of the line is
`3x+1=6y-2=1-z`
`rArr" "3(x+(1)/(3))=6(y-(1)/(3))=-(z-1)`
`rArr" "(x+1//3)/(1//3)=(y-1//3)/(1//6)=(z-1)/(-1)`
`therefore ` The line is passing through the point `A(-1//3. 1//3,1)` and having direction 1/3, 1/6. -1, i.e., 2, 1, -6.
Let `veca` be the position through the point A w. r. t. the origin and `vecb` be the vector parallel to the line
`therefore veca = -1//3 hati + 1//3 hatj +hatk and vecb = 2hati + hatj - 6 hatk`
`therefore ` The vector form of the equation of the line is
`vecr = vec a + lambda vecb`
`vecr = (-1//3 hati + 1//3 hatj + hatk) + lambda ( 2hati + hatj- 6hatk)`