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Question: The equation of straight line passing through the point (a, b, c) and parallel to z- axis, is...

The equation of straight line passing through the point (a, b, c) and parallel to z- axis, is

A

x−a1=y−b1=z−c0\frac { x - a } { 1 } = \frac { y - b } { 1 } = \frac { z - c } { 0 }

B

x−a0=y−b1=z−c1\frac { x - a } { 0 } = \frac { y - b } { 1 } = \frac { z - c } { 1 }

C

x−a1=y−b0=z−c0\frac { x - a } { 1 } = \frac { y - b } { 0 } = \frac { z - c } { 0 }

D

x−a0=y−b0=z−c1\frac { x - a } { 0 } = \frac { y - b } { 0 } = \frac { z - c } { 1 }

Answer

x−a0=y−b0=z−c1\frac { x - a } { 0 } = \frac { y - b } { 0 } = \frac { z - c } { 1 }

Explanation

Solution

The line through (a,b,c)( a , b , c ) is x−al=y−bm=z−cn\frac { x - a } { l } = \frac { y - b } { m } = \frac { z - c } { n } …..(i)

Since the line is parallel to z-axis, therefore any point on this line will be of the form (a,b,z1)\left( a , b , z _ { 1 } \right)

Also any point on line (i) is (lr+a,mr+b,nr+c)( l r + a , m r + b , n r + c )

Hence

Hence the line will be x−a0=y−b0=z−c1\frac { x - a } { 0 } = \frac { y - b } { 0 } = \frac { z - c } { 1 }.