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Question

The equation of the plane containing the line 2x - 5y + z = 3, x + y + 4z = 5 and parallel to the plane x + 3y + 6z = 1, is 
  1. 2x + 6y + 12z = -13
  2. 2x + 6y + 12z = 13
  3. x + 3y + 6z = -7
  4. x + 3y + 6z = 7


Solution

The correct option is D x + 3y + 6z = 7
Equation of any plane which passes through the intersection of the planes 2x - 5y + z = 3 and  x + y + 4z = 5 is given by 
2x5y+z3+λ(x+y+4z5)=0
x(2+λ)+y(λ5)+z(4λ+1)35λ=0
We are given this plane is parallel to x + 3y + 6z = 1. Comparing the coefficients, we get 
λ+21=λ53=4λ+16
3λ+6=λ5
2λ=11 
λ=112
We will now substitute this value in the equation of plane.
 equation of plane 7x221y221z+492=0          7x+21y+42z49=0
x+3y+6z7=0

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