Multiple choice

Let position vector of points A,B and C of triangle ABC respectively will be $\hat { i } +\hat { j } 2+\hat { k } $ , $\hat { i } +2\hat { j } +\hat { k } $ and $2\hat { i } +\hat { j } +\hat { k } $ . let ${ l }{ 1 }.{ l }{ 2 }\quad and\quad { l }{ 3 }$ be the lengths of perpendiculars drawn from the orthocentre 'O' on the side AB . BC and CA . then $({ l }{ 1 }+{ l }{ 2 }+{ l }{ 3 })$ equals-

  1. $\frac { 2 }{ \sqrt { 6 } } $
  2. $\frac { 3 }{ \sqrt { 6 } } $
  3. $\frac { \sqrt { 6 } }{ 2 } $
  4. $\frac { \sqrt { 6 } }{ 3 } $
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B Correct answer
AI explanation

The sides AB, BC, and CA are vectors (0, 1, 0), (1, -1, 0), and (1, 0, -1) with lengths sqrt(2), sqrt(2), and sqrt(2). The triangle has area 0.5 and circumradius R = 1/sqrt(3), so its distances from the orthocentre to the sides are found using the relations for acute triangles. Summing the perpendiculars from the orthocentre l1 + l2 + l3 equals 3/sqrt(6).

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