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JEE MainMathematicsVector Algebra

Let ABC be a triangle. Points D, E, and F lie on the sides BC, CA, and AB respectively, and divide them internally in the ratios 3:1 , 1:3 , and 3:1 . If AD + BE + CF = k AC , then the value of k is

Options

  1. A- 1 2
  2. B0
  3. C1 2
  4. D2

Correct answer

C. 1 2

Step-by-step solution

Let the position vectors of the vertices A, B, and C be a , b , and c respectively. The point D divides BC in the ratio 3:1 . By the section formula, the position vector of D is: d = 1 b + 3 c 3+1 = b + 3 c 4 Similarly, E divides CA in the ratio 1:3 , so its position vector is: e = 3 c + 1 a 1+3 = 3 c + a 4 And F divides AB in the ratio 3:1 , so its position vector is: f = 1 a + 3 b 3+1 = a + 3 b 4 Now, we find the vectors AD , BE , and CF : AD = d - a = b + 3 c - 4 a 4 BE = e - b = 3 c + a - 4 b 4 CF = f - c = a +

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