MHTCET 5 May 2025 Shift 2 Paper

Concept:Use the scalar triple product property and the fact that $\bar{a}$ and $\bar{b}$ are perpendicular unit vectors.Explanation:First verify the given properties.$|\bar{a}|^2 = \frac{1}{10}(9+0+1) = 1$, so $|\bar{a}| = 1$.$|\bar{b}|^2 = \frac{1}{49}(4+9+36) = 1$, so $|\bar{b}| = 1$.Also, $\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(6 - 6) = 0$.Therefore, $\bar{a}$ and $\bar{b}$ are perpendicular unit vectors, so $|\bar{a} \times \bar{b}| = 1$.Rewrite the given expression as a scalar triple product:$$(\bar{a}-2\bar{b}) \cdot \{(\bar{a} \times \bar{b}) \times (2\bar{a}+\bar{b})\} = [\bar{a}-2\bar{b}, \bar{a} \times \bar{b}, 2\bar{a}+\bar{b}]$$Interchanging the first two vectors changes the sign:$$= -(\bar{a} \times \bar{b}) \cdot \{(\bar{a}-2\bar{b}) \times (2\bar{a}+\bar{b})\}$$Now simplify the cross product:$$(\bar{a}-2\bar{b}) \times (2\bar{a}+\bar{b}) = \bar{a}\times\bar{b} + 4(\bar{a}\times\bar{b}) = 5(\bar{a}\times\bar{b})$$Substitute this back:$$-(\bar{a} \times \bar{b}) \cdot \{5(\bar{a} \times \bar{b})\} = -5|\bar{a} \times \bar{b}|^2 = -5(1)^2 = -5$$Answer:The value is $-5$, so the correct option is B.