The equation of a line given by `(4-x)/3=(y+3)/3=(z+2)/6dot` Write the direction cosines of a line parallel to this line.

Explanation

We have given the equation of a line as `(4-x)/3=(y+3)/3=(z+2)/6`

It can be written as

`(x-4)/(-3)=(y+3)/3=(z+2)/6`
Now the direction ratios are (-3,3,6)

Therefore direction cosines of the line parallel to this line are

`(-3)/(sqrt((-3)^2+3^2+6^2)),(3)/(sqrt((-3)^2+3^2+6^2)),(6)/(sqrt((-3)^2+3^2+6^2))`

`=(-1)/(sqrt6),1/(sqrt6),2/(sqrt6)`