Question

plz do answer this.......

Answer 1

in cartesian form above equations can be written as:

P1: x+y+z-1=0

P2: 2x+3y-z+4=0

therefore, eq. of a plane passing through the intersection of the above planes is

(x+y+z-1) + u(2x+3y-z+4)=0

or, (1+2u)x +(1+3u)y + (1-u)z -1+4u=0

as the plane is parallel to x-axis

therefore,

(1+2u).1=0

u=(-1)/2

$(1 + 2. \frac{( - 1)}{2} )x + (1 + 3 \frac{( - 1)}{2} )y + (1 - \frac{ ( - 1)}{2} )z - 1 + 4( \frac{( - 1)}{2} ) = 0 \\ \frac{( - 1)}{2} y + \frac{3}{2} z - 1 - 2 = 0 \\ \frac{ - y}{2} + \frac{3z}{2} - 3 = 0 \\ on \: multipling \: both \: sides \: by \: ( - 2) \: we \: get \\ y - 3z + 6 = 0$