Q. 244.8( 6 Votes )

Find the equation

Answer :

Given: equation of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 =0


To find: the equation of the plane passing through the line of intersection of the two given and parallel to the line


Equations of the planes are


2x + y – z -3=0 and 5x – 3y + 4z + 9 =0


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


2x + y – z - 3 +λ(5x – 3y + 4z + 9)=0


2x + y – z - 3 +5λx-3λy+4z λ+9λ=0


(2+5λ)x+(1-3λ)y+(4λ-1)z+(9λ-3)=0……………(iii)


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


a1=2+5λ, b1=1-3λ, c1=4λ-1


Now plane (iii) is parallel to line



Now the direction ratios of this line are,


a2=2, b2=4, c2=5


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


a1a2+b1b2+ c1c2=0


Substituting the corresponding values, we get


(2+5λ)(2)+(1-3λ)(4)+( 4λ-1)(5)=0


4+10λ+4-12 λ+20 λ-5=0


3+18λ=0


18λ=-3



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







Multiplying both sides with 6 we get


7x+9y-10z-27=0


7x+9y-10z=27


Is the required equation of the plane


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