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Let O and H be the circumcentre and the orthocentre of a ABC , respectively. If the circumradius of the triangle is 3 and the distance between the circumcentre and the orthocentre is 15 , then the value of OA OB + OB OC + OC OA is

Options

  1. A-6
  2. B6
  3. C-12
  4. D3

Correct answer

A. -6

Step-by-step solution

Let the circumcentre O be the origin. Since the circumradius is 3 , the magnitudes of the position vectors of the vertices are | OA | = | OB | = | OC | = 3 . The position vector of the orthocentre H with respect to the circumcentre O is given by OH = OA + OB + OC . We are given that the distance OH = 15 , so | OH |^2 = 15 . Squaring the vector equation for OH , we get: | OH |^2 = | OA + OB + OC |^2 15 = | OA |^2 + | OB |^2 + | OC |^2 + 2( OA OB + OB OC + OC OA ) Substituting the magnitudes of the position vectors:

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