Question

Find the direction cosines of the line `(2x - 1)/3 = 3y = (4z + 3)/2`

Answer 1

`(2x - 1)/3 = 3y = (4z + 3)/2`

The equations can be written as

`(x - 1/2)/(3/2) = y/(1/3) = (z + 3/4)/(2/4)`

These equations are of the form

`(x - x_1)/a = (y - y_1)/b = (z - z_1)/c`

∴ Direction ratios of the line are

`3/2, 1/3, 1/2`

i.e. 9, 2, 3

∴ `sqrt((9)^2 + (2)^2 + (3)^2)`

= `sqrt(81 + 4 + 9)`

= `sqrt(94)`

∴ d.c.s. of the line are `9/sqrt(94), 2/sqrt(94), 3/sqrt(94)`