The equation of a line `4x-4y-z+11=0=x+2y-z-1` can be put as

Options

A. ` ( x ) / (2 ) = ( y - 2 ) / ( 1) = ( z - 3 ) / ( 4 ) `
B. ` ( x - 4 ) / ( - 2 ) = ( y - 4 ) /( 2 ) = ( z + 11 ) / ( 2 ) `
C. ` ( x - 2 ) / (2 ) = ( y ) / ( 1 ) = ( z - 3 ) / (4 ) `
D. ` ( x- 2 ) / ( 2 ) = ( y - 2 ) / ( 1 ) = ( z ) / ( 4 ) `

Answer

A

Explanation

The given equaions are
` 4x - 4y - z + 11 = 0" " `…(i)
and ` x + 2y - z - 1 = 0 " " `…(ii)
The DR's of normals to the planes (i) and (ii) are 4, - 4, - 1 and 1, 2, - 1 respectively.
Let DR's of the line intersection of plane be l, m, n.
As the line of intersection of the planes is perpendicular to the normals of the both planes, we get
` 4l - 4m - n = 0 `
and ` l + 2m - n = 0 `
By cross multiplication
` ( l ) / ( 6 ) = ( m ) / (3 ) = ( n ) / ( 3 ) or ( l ) / (2 ) = ( m ) / (1 ) = ( n ) / ( 4 ) `
If x = 0 , Eqs. (i) and (ii) become
` - 4y - z + 11 = 0 or 2y- z - 1 = 0 `
Solving , we get ` y = 2 , z = 3 `
` therefore ` Equation of line is ` ( x ) / ( 2 ) = ( y - 2 ) / ( 1 ) = ( z - 3 ) / ( 4 ) `