Find the equation of the plane through the points `A(2,1,-1)` and `B(-1,3,4)` and perpendicular to the plane `x-2y+4z=10`. Also, show that the plane thus obtained contains the line `vecr=(-hati+3hatj+4hatk)+lambda(3hati-2hatj-5hatk)`.

Answer

`18x+17y+4z=49`

Explanation

Obtain the required equation of the plane as `18x+17y+4z=49`………(A)
The given line is `vecr=(3lambda-1)hati+(3-2lambda)hatj+(4-5lambda)hatk`.
Coordinates of any points on this line are `(3lambda-1,3-2lambda,4-5lambda)`.
These coordinates satisfy (A), as shown below:
LHS =`18(3lambda-1)+17(3-2lambda)+4(4-5lambda)=49` =RHS.
So, `(3lambda-1,3-2lambda,4-5lambda)` lies on plane (A).
Hence, the plane (A) obtained above equation the given line.