Question

The vector eqliation of line 2x - 2 = 3y + 1 = 6z - 2 is

Options

• `bar"r" = (hat"i" - 2/3hat"j" + 2hat"k") + lambda(hat"i" + 2hat"j" + 6hat"k")`

• `bar"r" = (hat"i" - 2/3hat"j" + 2/3hat"k") + lambda(3hat"i" + 2hat"j" + hat"k")`

• `bar"r" = (hat"i" - 1/3hat"j" + 1/3hat"k") + lambda(3hat"i" + 2hat"j" + hat"k")`

• `bar"r" = (hat"i" - 1/3hat"j" - 1/3hat"k") + lambda(3hat"i" + 2hat"j" + 6hat"k")`

Answer 1

`bar"r" = (hat"i" - 1/3hat"j" + 1/3hat"k") + lambda(3hat"i" + 2hat"j" + hat"k")`

Explanation:

Given cartesian equation of the line is

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

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

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

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

The given line passes through `(1, (-1)/3, 1/3)` and the direction ratios are proportional to 3, 2, 1.

∴ The vector equation is

`bar"r" = (hat"i" - 1/3hat"j" + 1/3hat"k") + lambda(3hat"i" + 2hat"j" + hat"k")`