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BITSAT Mathematics Vector Algebra 2024 BITSAT 2024 (Memory Based Paper 2)

BITSAT Mathematics Question (2024) — Solution

Question

Let ABC be a triangle and be a , b , c the position vectors of A , B , C respectively. Let D divide B C in the ratio 3: 1 internally and E divide AD in the ratio 4: 1 internally. Let BE meet AC in F . IfE divides BF in the ratio 3: 2 internally then the position vector of F is

Options

  1. A. a + b + c 3
  2. B. a -2 b +3 c 2
  3. C. a +2 b +3 c 2
  4. D. a - b +3 c 3

Answer

D. a - b +3 c 3

Step-by-step solution

Here we are given that OA = a , OB = b , OC = c Now P.V of Di.e OD = 1 OB +3 OC 1+3 aligned & OD = b +3 c 4 \\ & OE = 4 OD + OA 4+1 = 4( ~b +3 c ) 4 + a 5 \\ & OE = a + b +3 c 5 \\ & Now, OE = 2 OB +3 OF 2+3 \\ & OF = 5 OE -2 OB 3 \\ & OF = 5( a + b +3 c ) 5 -2 ~b 3 \\ & OF = a - b +3 c 3 P . V . of F is a - b +3 c 3 aligned

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