Mht CetVector AlgebraMHT CET 2026 18th April Evening ShiftMCQ+2 / -02026If $\bar{a} \cdot \bar{b} = \beta$ and $\bar{a} \times \bar{b} = \bar{c}$ then $\bar{a} = $A$\dfrac{\bar{b} \times \bar{c} - \beta\bar{b}}{|\bar{b}|^2}$B$\dfrac{\bar{b} \times \bar{c} - \beta\bar{c}}{|\bar{b}|^2}$C$\dfrac{\bar{b} \times \bar{c} + \beta\bar{b}}{|\bar{b}|^2}$D$\dfrac{\bar{b} \times \bar{c} + \beta\bar{c}}{|\bar{b}|^2}$Check AnswerClear SelectionReveal AnswerShow Explanation