Given: Plane passes through the line of intersection of the planes
2x+y−z=3,5x−3y+4z+9=0 and is parallel to the line
== As we know that, equation of plane passing through the line of intersection of two planes
a1x+b1y+c1z+d1=0 and
a2x+b2y+c2z+d2=0 is given by:
(a1x+b1y+c1z+d1) +λ(a2x+b2y+c2z+d2)=0 where
λ∈R.
So, the equation of plane passing through the line of intersection of the planes
2x+y−z=3,5x−3y+4z+9=0 is given by:
⇒(2x+y−z−3)+λ(5x−3y +4z+9)=0 ⇒(2+5λ)x+(1−3λ)y+(4λ−1)z +(9λ−3)=0 ..........(1)
∵ It is given that the plane which passes through the line of intersection of the planes x - 2y + z = 1 and 2x + y + z = 8 is parallel to the line
==. So, the normal to the plane represented by (1) is perpendicular to the line
== The direction ratios of the plane represented by (1) are:
2+5λ,1−3λ,4λ−1 Similarly, the direction ratios of the line
== are:
2,4,5.
As we know that if two lines are perpendicular then
a1.a2+b1.b2+c1.c2=0 where
a1,b1,c1,a2,b2 and
c2 are the direction ratios
⇒2(2+5λ)+4(1−3λ) +5(4λ−1)=0 ⇒λ=−1∕6 By substituting λ = - 1/6 in equation (1), we get
⇒7x+9y−10z−27=0 Hence, the required equation of plane is:
7x+9y−10z−27=0