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KCET2024MathematicsThree Dimensional Geometry

The plane containing the point (3,2,0) and the line x-3 1 = y-6 5 = z-4 4 is

Options

  1. Ax-y+z=1
  2. Bx+y+z=5
  3. Cx+2 y-z=1
  4. D2 x-y+z=5

Correct answer

A. x-y+z=1

Step-by-step solution

Given that point (3,2,0) lies on plane and line x-3 1 = y-6 5 = z-4 4 also lies on the plane. Plane contains points (3,2,0) and (3,6,4) The DR's of line joining these two points is (0,4,4) . Therefore, DR's of normal to plane is ( i +5 j +4 k ) (4 j +4 k )=4 i -4 j +4 k Therefore, the equation of plane is 4 x-4 y+4 z=k Point (3,2,0) lies on planes. So, we get k=4 3-4 2+4 0=4 Hence, the equation of plane is x-y+z=1 .

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