Q. 295.0( 1 Vote )

Find the equation

Answer :

Given: equation of the planes and


To find: the equation of the plane passing through the line of intersection of the two givenand parallel to the x-axis.


The given equation of planes can be written in the Cartesian form, as shown below




x+y+z-1=0……..(i)





2x+3y-z+4=0……..(ii)


Hence given equation of planes can be written in Cartesian form are


x+y+z-1=0


2x+3y-z+4=0


The equation of a plane passing through the line of intersection of these two planes is,


x+y+z-1+λ(2x+3y-z+4)=0


x+y+z-1+2 λx+3 λy-z λ+4λ=0


(1+2λ)x+(1+3λ)y+(1-λ)z+(4λ-1)=0……………(iii)


Now, direction ratios of normal to the plane (iii), are,


a1=1+2λ, b1=1+3λ, c1=1-λ


Now the direction ratios of x-axis are,


a2=1, b2=0, c2=0


Since the required plane (iii) is parallel to x-axis, then


a1a2+b1b2+ c1c2=0


Substituting the corresponding values, we get


(1+2λ)(1)+(1+3λ)(0)+(1-λ)(0)=0


1+2λ+0+0=0


2λ=-1



So substituting the value of λ in equation (iii), we get






-y+3z-6=0


y-3z+6=0


Is the required equation of the plane, it can also be written as



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