MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use the vector triple product identity and compare magnitudes to find the angle.Explanation:By the vector triple product identity:$$\bar{a} \times (\bar{a} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{a} - (\bar{a} \cdot \bar{a})\bar{c}$$Since $|\bar{a}| = 1$, we have $\bar{a} \cdot \bar{a} = 1$.Thus,$$\bar{a} \times (\bar{a} \times \bar{c}) = (\bar{a} \cdot \bar{c})\bar{a} - \bar{c}$$Given $\bar{a} \times (\bar{a} \times \bar{c}) + \bar{b} = \overline{0}$, we get:$$\bar{b} = -\bar{a} \times (\bar{a} \times \bar{c})$$Taking magnitudes on both sides:$$|\bar{b}| = |\bar{a} \times (\bar{a} \times \bar{c})|$$Since $|\bar{b}| = 1$,$$|\bar{a} \times (\bar{a} \times \bar{c})| = 1$$Now, $\bar{a} \times (\bar{a} \times \bar{c})$ has magnitude $|\bar{c}|\sin\theta$, where $\theta$ is the angle between $\bar{a}$ and $\bar{c}$.Given $|\bar{c}| = 2$, we have:$$2\sin\theta = 1$$$$\sin\theta = \frac{1}{2}$$The acute angle satisfying this is:$$\theta = \frac{\pi}{6}$$Answer:The acute angle between $\bar{a}$ and $\bar{c}$ is $\frac{\pi}{6}$, which corresponds to option D.