# Find the Vector Equation of the Plane Which Contains the Line of Intersection of the Planes - Mathematics

Find the vector equation of the plane which contains the line of intersection of the planes vecr (hati+2hatj+3hatk)-4=0 and vec r (2hati+hatj-hatk)+5=0 which is perpendicular to the plane.vecr(5hati+3hatj-6hatk)+8=0

#### Solution

The equations of the given planes are

vecr (hati+2hatj+3hatk)-4=0 ........(1)

vec r (2hati+hatj-hatk)+5=0  .........(2)

The equation of the plane passing through the intersection of the planes (1) and (2) is

[vecr (hati+2hatj+3hatk)-4]+lambda[ vec r (2hati+hatj-hatk)+5]=0

vecr[(1+2lambda)hati+(2+lambda)hatj+(3-lambda)hatk]=4-5lambda.............(3)

Given, the plane (3) is perpendicular to the plane vecr(5 hati+3hatj-6hatk)+8=0

(1+2lambda)xx5+(2+lambda)xx3+(3-lambda)xx(-6)=0

19lambda-7=0

lambda=7/19

Putting lambda=7/19  in (3), we get

vec r[(1+14/19)hati+(2+7/19)hatj+(3-7/19)hatk]=4-35/19

vecr(33/19hati+45/19hatj+50/19hatk)=41/19

vecr(33hati+45hatj+50hatk)=41

Thus, the equation of the required plane is

vecr(33hati+45hatj+50hatk).=41

Concept: Vector and Cartesian Equation of a Plane
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