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KCET2016MathematicsThree Dimensional Geometry

The vector equation of the plane which is at a distance of ( 3 14 ) from the origin and the normal from the origin is ( 2 i -3 j + k ) is

Options

  1. A( r (2 i -3 j + k )=3 )
  2. B( r ( i + j + k )=9 )
  3. C( r ( i +2 j )=3 )
  4. D( r (2 i + k )=3 )

Correct answer

A. ( r (2 i -3 j + k )=3 )

Step-by-step solution

Given that ( N =2 i -3 j + k ) So, ( | N |= 4+9+1 = 14 ) Unit vector is given by ( n = N 14 ) and from origin is ( d= 3 14 ) So, required equation, ( r n =d ) ( r 2 i -3 j + k = 3 14 ) ( r (2 i -3 j + k )=3 )

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