Find the direction cosines of teh line `(x-2)/2=(2y-5)/(-3),z=-1.` Also, find the vector equation of the line.

Options

A. `(4)/(3),(3)/(5),0`
B. `(3)/(5),(4)/(5),(1)/(5)`
C. `-(3)/(5),(4)/(5),0`
D. `(4)/(5),-(2)/(5),(1)/(5)`

Answer

A

Explanation

Given equation of line is `(x+2)/(2)=(2y-5)/(3), z+1=0`
`or (x+2)/(2)=((y-(5)/(2))/((3)/(2))), z+1=0`
Hence, DR of a line are `lt 2, (3)/(2), 0gt`
Now, `sqrt(2^(2)+((3)/(2))^(2)+0)), sqrt(4+(9)/(4))=sqrt((25)/(4)))=(5)/(2)`
`therefore `DC of a line are `lt (2)/(5//2), (3//2)/(5//2), 0 gt or (4)/(5), (3)/(5), 0 gt`