# Find the equation of the plane passing through the line of intersection of planes 2x - y + z = 3 and 4x- 3y + 5z + 9 = 0 and parallel to the line (x+1)/2=(y+3)/4=(z-3)/5 - Mathematics and Statistics

Find the equation of the plane passing through the line of intersection of planes 2x - y + z = 3 and 4x- 3y + 5z + 9 = 0 and parallel to the line

 (x+1)/2=(y+3)/4=(z-3)/5

#### Solution

Given planes are 2x-y+z =3, 4x-3y+5z+9 =0
Equation of required plane passing through their intersection is
(2x - y + z -3) +λ(4x-3y+5z + 9)= 0 .....(1)
(2 + 4λ) x +(-1-3λ) y+(1+5λ) z+(-3+ 9λ)= 0
Directio ratios of the normal to the above plane are 2+ 4λ,  -1-3λ and 1+ 5λ

Plane is parallel to the line  (x+1)/2=(y+3)/4=(z-3)/5

Direction ratios of line are 2, 4,5
Given that required plane is parallel to given line.
∴normal of the plane is perpendicular to the given line

(2+4lambda)2+(-1-3lambda)4+(1+5lambda)5=0

4+8lambda-4-12lambda+5+25 lambda=0

21lambda+5=0

therefore lambda=-5/21

Substituting λ in (1)
∴ Equation of plane is

(2x-y+z-3)-5/21(4x-3y+5z+9)=0

42x-21y+21z-63-20x+15y-25z-45=0

22x-6y-4z-108=0

11x-3y-2z-54=0

Concept: Plane - Equation of Plane Passing Through the Intersection of Two Given Planes
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2013-2014 (March)

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