MHT CET · Physics · Mathematics in Physics
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
- 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 \(\frac{a_{1}}{b_{2}}=-\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}\)
Answer & Solution
Correct Answer
(D) \(\frac{a_{1}}{b_{2}}=-\frac{\mathrm{a}_{2}}{\mathrm{~b}_{1}}\)
Step-by-step Solution
Detailed explanation
\( \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}} = 0 \) \( a_{1}b_{1} + a_{2}b_{2} = 0 \)
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