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Manipal MET2016MathematicsThree Dimensional Geometry

The vector equation of the plane through the point i +2 j - k and perpendicular to the line of intersection of the plane r (3 i - j + k )=1 and r ( i -4 j +2 k )=2 is

Options

  1. Ar (2 i +7 j -13 k )=1
  2. Br (2 i -7 j -13 k )=1
  3. Cr (2 i +7 j +13 k )=0
  4. DNone of the above

Correct answer

B. r (2 i -7 j -13 k )=1

Step-by-step solution

The line of intersection of the planes r (3 i - j + k )=1 and r ( i +4 j -2 k )=2 is common to both the planes. Therefore, it is perpendicular to normals to the two planes i.e. n ₁=3 i - j + k and n ₂= i +4 j -2 k . Hence, it is parallel to the vector n ₁ n ₂=-2 i +7 j +13 k Thus, we have to find the equation of the plane passing through a = i +2 j - k and normal to the vector n = n ₁ n ₂ . The equation of the required plane is ( r - a ) n or r n = a n or r (-2 i +7 j +13 k )=( i +2 i + k ) (-2 i +7 j +13 k ) or r

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