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AP EAMCET · Maths · Vector Algebra

For two given vectors \(\bar{a}\) and \(\bar{b}\) if the vectors \(\overline{\mathrm{A}}, \overline{\mathrm{B}}\) are such that \(\overline{\mathrm{A}}+\overline{\mathrm{B}}=\bar{a}, \overline{\mathrm{A}} \times \overline{\mathrm{B}}=\bar{b}\) and \(\overline{\mathrm{A}} \cdot \bar{a}=1\), then \(\overline{\mathrm{A}}=\)

  1. A \(\frac{(\bar{a} \times \bar{b})+\bar{a}}{\bar{a}^2}\)
  2. B \(\frac{(\bar{b} \times \bar{a})+\bar{a}\left(\bar{a}^2-1\right)}{\bar{a}^2}\)
  3. C \(\frac{\bar{a}\left(\bar{a}^2-1\right)+\bar{b}\left(\bar{b}^2-1\right)}{\bar{a}^2+\bar{b}^2}\)
  4. D \(\frac{(\bar{a} \times \bar{b})+\bar{b}}{\bar{b}^2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{(\bar{a} \times \bar{b})+\bar{a}}{\bar{a}^2}\)

Step-by-step Solution

Detailed explanation

No solution. Refer to answer key.