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JEE Main 2025 (23 Jan, Shift 1) · Scalar (Dot) Product

Scalar (Dot) Product question from JEE Main 2025 (23 Jan, Shift 1)

Q15JEE Main 2025 · 23 JAN · SHIFT 1Medium

Scalar (Dot) Product

Two particles are located at equal distance from origin. The position vectors of those are represented by A⃗=2i^+3nj^+2k^\vec{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k} and B⃗=2i^−2j^+4pk^\vec{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}, respectively. If both the vectors are at right angle to each other, the value of n−1n^{-1} is ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Equal distance: 4+9n2+4=4+4+16p2⇒9n2=16p2, p=±3n44 + 9n^2 + 4 = 4 + 4 + 16p^2 \Rightarrow 9n^2 = 16p^2,\ p = \pm\frac{3n}{4}.
  2. Perpendicular: A⃗⋅B⃗=4−6n+8p=0\vec{A} \cdot \vec{B} = 4 - 6n + 8p = 0.
  3. p=+3n4p = +\frac{3n}{4} gives 4=04 = 0 (impossible); p=−3n4p = -\frac{3n}{4} gives 4−12n=0⇒n=134 - 12n = 0 \Rightarrow n = \frac{1}{3}.
  4. Answer: n−1=3n^{-1} = 3
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