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Question: Answered & Verified by Expert
If $\overrightarrow{\mathrm{A}}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}$ and $\overrightarrow{\mathrm{B}}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}$ are perpendicular to each other then
PhysicsMathematics in PhysicsMHT CETMHT CET 2020 (15 Oct Shift 2)
Options:
  • A $\frac{\mathrm{b}_{2}}{\mathrm{a}_{1}}=-\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}$
  • B $\frac{\mathrm{a}_{1}}{\mathrm{~b}_{2}}=+\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}$
  • C $\frac{\mathrm{b}_{2}}{\mathrm{a}_{1}}=+\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}$
  • D $\frac{a_{1}}{b_{2}}=-\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}$
Solution:
2654 Upvotes Verified Answer
The correct answer is: $\frac{a_{1}}{b_{2}}=-\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}$
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