Solveeit Logo
Login

Question

Question: The d.c.’s of the line \(6 x - 2 = 3 y + 1 = 2 z - 2\) are...

The d.c.’s of the line 6x−2=3y+1=2z−26 x - 2 = 3 y + 1 = 2 z - 2 are

A

13,13,13\frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } } , \frac { 1 } { \sqrt { 3 } }

B

114,214,314\frac { 1 } { \sqrt { 14 } } , \frac { 2 } { \sqrt { 14 } } , \frac { 3 } { \sqrt { 14 } }

C

1, 2, 3

D

None of these

Answer

114,214,314\frac { 1 } { \sqrt { 14 } } , \frac { 2 } { \sqrt { 14 } } , \frac { 3 } { \sqrt { 14 } }

Explanation

Solution

We have 6x−2=3y+1=2z−26 x - 2 = 3 y + 1 = 2 z - 2

⇒ 6x−(2/6)1=3y+(1/3)1=2(z−1)1\frac { 6 x - ( 2 / 6 ) } { 1 } = \frac { 3 y + ( 1 / 3 ) } { 1 } = \frac { 2 ( z - 1 ) } { 1 }

⇒ x−(1/3)1/6=y+(1/3)1/3=z−11/2\frac { x - ( 1 / 3 ) } { 1 / 6 } = \frac { y + ( 1 / 3 ) } { 1 / 3 } = \frac { z - 1 } { 1 / 2 }

⇒ x−(1/3)1=y+(1/3)2=z−13\frac { x - ( 1 / 3 ) } { 1 } = \frac { y + ( 1 / 3 ) } { 2 } = \frac { z - 1 } { 3 }

d.r.’s of line are (1, 2, 3).

Hence d.c.’s of line are (1/14,2/14,3/14)( 1 / \sqrt { 14 } , 2 / \sqrt { 14 } , 3 / \sqrt { 14 } )