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JEE Main202427 Jan 2024Evening ShiftMathematicsThree Dimensional GeometryActual

Let the position vectors of the vertices A , B and C of a triangle be 2 i ^ + 2 j ^ + k ^ , i ^ + 2 j ^ + 2 k ^ and 2 i ^ + j ^ + 2 k ^ respectively. Let l 1 , l 2 and l 3 be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides AB , BC and CA respectively, then l 1 2 + l 2 2 + l 3 2 equals :

Options

  1. A1 5
  2. B1 2
  3. C1 4
  4. D1 3

Correct answer

B. 1 2

Step-by-step solution

Given: A 2 , 2 , 1 , B 1 , 2 , 2 and C 2 , 1 , 2 are vertices of ∆ ABC . ⇒ AB = 2 - 1 2 + 2 - 2 2 + 1 - 2 2 = 2 ⇒ BC = 1 - 2 2 + 2 - 1 2 + 2 - 2 2 = 2 ⇒ CA = 2 - 2 2 + 2 - 1 2 + 1 - 2 2 = 2 So, ∆ ABC is equilateral. Therefore, orthocentre and centroid will be same. ⇒ G ≡ 2 + 1 + 2 3 , 2 + 2 + 1 3 , 1 + 2 + 2 3 ⇒ G ≡ 5 3 , 5 3 , 5 3 Also, mid point of AB is D 3 2 , 2 , 3 2 ⇒ l 1 = 3 2 - 5 3 2 + 2 - 5 3 2 + 3 2 - 5 3 2 ⇒ l 1 = 1 36 + 1 9 + 1 36 ⇒ l 1 = 1 6 Similarly, l 2 = l 3 = 1 6 ⇒ l 1 2 + l 2 2 + l 3 2 = 1 6 + 1

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