Q71 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
If $\bar{a} \cdot \bar{b} = \beta$ 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$\dfrac{\bar{b} \times \bar{c} - \beta\bar{c}}{|\bar{b}|^2}$
- Correct answerC$\dfrac{\bar{b} \times \bar{c} + \beta\bar{b}}{|\bar{b}|^2}$
- D$\dfrac{\bar{b} \times \bar{c} + \beta\bar{c}}{|\bar{b}|^2}$
Solution
Given $\bar{a} \times \bar{b} = \bar{c}$
Taking cross product with $\bar{b}$ on both sides from the left:
$\bar{b} \times (\bar{a} \times \bar{b}) = \bar{b} \times \bar{c}$
Using the vector triple product expansion $\bar{x} \times (\bar{y} \times \bar{z}) = (\bar{x} \cdot \bar{z})\bar{y} - (\bar{x} \cdot \bar{y})\bar{z}$:
$(\bar{b} \cdot \bar{b})\bar{a} - (\bar{b} \cdot \bar{a})\bar{b} = \bar{b} \times \bar{c}$
Substituting $\bar{b} \cdot \bar{b} = |\bar{b}|^2$ and $\bar{a} \cdot \bar{b} = \beta$:
$|\bar{b}|^2 \bar{a} - \beta \bar{b} = \bar{b} \times \bar{c}$
$|\bar{b}|^2 \bar{a} = \bar{b} \times \bar{c} + \beta \bar{b}$
$\bar{a} = \dfrac{\bar{b} \times \bar{c} + \beta \bar{b}}{|\bar{b}|^2}$
Taking cross product with $\bar{b}$ on both sides from the left:
$\bar{b} \times (\bar{a} \times \bar{b}) = \bar{b} \times \bar{c}$
Using the vector triple product expansion $\bar{x} \times (\bar{y} \times \bar{z}) = (\bar{x} \cdot \bar{z})\bar{y} - (\bar{x} \cdot \bar{y})\bar{z}$:
$(\bar{b} \cdot \bar{b})\bar{a} - (\bar{b} \cdot \bar{a})\bar{b} = \bar{b} \times \bar{c}$
Substituting $\bar{b} \cdot \bar{b} = |\bar{b}|^2$ and $\bar{a} \cdot \bar{b} = \beta$:
$|\bar{b}|^2 \bar{a} - \beta \bar{b} = \bar{b} \times \bar{c}$
$|\bar{b}|^2 \bar{a} = \bar{b} \times \bar{c} + \beta \bar{b}$
$\bar{a} = \dfrac{\bar{b} \times \bar{c} + \beta \bar{b}}{|\bar{b}|^2}$
Hence, the correct answer is option: 'C'.