Direction cosines of the line `(x+2)/(2)=(2y-5)/(3),z=-1` are
A
`4/5,3/5,0`
B
`3/5,4/5,1/5`
C
`-(3)/(5),(4)/(5),0`
D
`4/5,-2/5,1/5`
Text Solution
Verified by Experts
The correct Answer is:
A
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's 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's of the lines are `lt (2)/(5//2), (3//2)/(5//2), 0 gt or lt 4/5, 3/5, 0 gt `
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