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

Physics · Vectors and Scalars

JEE Main 2025 — 23 January, Morning Shift — Question 64

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

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0$

$4-6 n+8 p=0$

$|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|$

$4+9 \mathrm{n}^{2}+4=4+4+16 \mathrm{p}^{2}$

$9 n^{2}=16 p^{2}$

$P=+\frac{3}{4} n$

$4-6 n \pm 6 n=0$

$12 \mathrm{n}=4$

$\mathrm{n}=\frac{1}{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Vectors and Scalars
Topic: Product of Vectors and Applications