Question

Find the vector and Cartesian equations of the line passing through the point (–1, –1, 2) and parallel to the line 2x − 2 = 3y + 1 = 6z − 2.

Answer 1

Let `bar"a"` be the position vector of the point A(–1, –1, 2) w.r.t. the origin.

Then `bar"a" = -hat"i" - hat"j" + 2hat"k"`

The equation of given 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)`

The direction ratios of this line are

`(1)/(2),(1)/(3),(1)/(6)` i.e. 3, 2, 1

Let `bar"b"` be the vector parallel to this line.

Then `bar"b" = 3hat"i" + 2hat"j" + hat"k"`

The vector equation of the line passing through `"A"(bara) "and parallel to"  bar"b"` is

`bar"r" = bar"a" + lambdabar"b"`, where λ is a scalar,

∴ the vector equation of the required line is

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

The line passes through (–1, – 1, 2) and has direction ratios 3, 2, 1

∴ the cartesian equations of the line are

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

i.e. `(x + 1)/(3) = (y + 1)/(2) = (z - 2)/(1)`.