Write the direction cosine of the line whose Cartesian equations are `2x=3y=-zdot`

Explanation

Given: `2x=3y=-z`

Then

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

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

Then direction ratios are `3,2-6`

Then direction cosines are `=3/(sqrt(3^2+2^2+(-6)^2)`,`2/(sqrt(3^2+2^2+(-6)^2)`,`(-6)/(sqrt(3^2+2^2+(-6)^2)`