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JEE Advanced Mathematics Vector Algebra 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Mathematics Question (2022) — Solution

Question

Let i ^ , j ^ and k ^ be the unit vectors along the three positive coordinate axes. Let a → = 3 i ^ + j ^ - k ^ b → = i ^ + b 2 j ^ + b 3 k ^ ,   b 2 , b 3 ∈ ℝ c → = c 1 i ^ + c 2 j ^ + c 3 k ^ , c 1 , c 2 , c 3 ∈ ℝ be three vectors such that b 2 b 3 > 0 , a → · b → = 0  and  0 - c 3 c 2 c 3 0 - c 1 - c 2 c 1 0 1 b 2 b 3 = 3 - c 1 1 - c 2 - 1 - c 3 .  Then, which of the following is/are TRUE?

Options

  1. A. a → · c → = 0
  2. B. b → · c → - = 0
  3. C. b → > 10
  4. D. c → ≤ 11

Answer

D. c → ≤ 11

Step-by-step solution

Given, a → = 3 i ^ + j ^ - k ^ b → = i ^ + b 2 j ^ + b 3 k ^ c → = c 1 i ^ + c 2 j ^ + c 3 k ^ 0 - c 3 c 2 c 3 0 - c 1 - c 2 c 1 0 1 b 2 b 3 = 3 - c 1 1 - c 2 - 1 - c 3 Now multiply and compare we get, b 2 c 3 - b 3 c 2 = c 1 - 3   . . . . . . . 1 c 3 - b 3 c 1 = 1 - c 2 . . . . . . . . . . . 2 c 2 - b 2 c 1 = 1 + c 3 . . . . . . . . . . . 3 Now applying operation  equation  1 i ^ - 2 j ^ + 3 k ^  we get, i ^ b 2 c 3 - c 2 b 3 - j ^ c 3 - b 3 c 1 + k ^ c 2 - b 2 c 1 = c 1 i ^ + c 2 j ^ + c 3 k ^ - 3 i ^ - j ^ + k ^ ⇒ b → × c → = c → - a →             . . . . i ⇒ b → × c → · b → = c → · b → - a → · b → ⇒ b → · c → = 0   given  a → . b → =0 Again from (i) c → × b → · c → = a → · c → - c → 2 = 0 ⇒ c → 2 = a → c → cos θ , where θ = a → ∧ c → ⇒ c → ≥ a → ⇒ c → ≥ 11 Given that a → · b → = 0 ⇒ b 2 - b 3 + 3 = 0 ⇒ b 3 - b 2 = 3 Also b 2 - b 3 > 0 Now b → = 1 + b 2 2 + b 3 2 = 1 + b 3 - b 2 2 + 2 b 2 b 3 = 10 + 2 b 2 b 3 ⇒ b → 2 > 10 ⇒ b → > 10

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Related: Mathematics — Vector Algebra · All PYQ Banks