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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

The equation of the plane through the points 2 , - 1,0 , 3 , - 4,5 and parallel to a line with direction cosines proportional to 2,3 and 4 is 9 x - 2 y - 3 z = k , then the value of k is

Options

  1. A20
  2. B- 20
  3. C10
  4. D- 10

Correct answer

A. 20

Step-by-step solution

Equation of the plane through 2 , - 1,0 is a x - 2 + b y + 1 + c z - 0 = 0 It also passes through 3 , - 4,5 , then a - 3 b + 5 c = 0 ... (i) Given plane is parallel to a line with direction cosines proportional to 2,3 , 4 , so 2 a + 3 b + 4 c = 0 ... (ii) Solving equations (i) and (ii), a - 12 - 15 = b 10 - 4 = c 3 + 6 a - 27 = b 6 = c 9 a - 9 = b 2 = c 3 = λ a = - 9 λ , b = 2 λ , c = 3 λ So, the equation of the plane is - 9 λ x - 2 + 2 λ y + 1 + 3 λ z - 0 = 0 - 9 x + 18 + 2 y + 2 + 3 z = 0 - 9 x + 2 y + 3 z = - 20

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