MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use the standard formulas for dot product and cross product magnitudes.Explanation:Let $|\bar{a}| = a$ and $|\bar{b}| = b$, with $\theta$ as the angle between them.The dot product is given by:$$|\bar{a} \cdot \bar{b}| = |ab \cos\theta|$$So, squaring it gives:$$|\bar{a} \cdot \bar{b}|^{2} = a^{2}b^{2}\cos^{2}\theta$$The magnitude of the cross product is:$$|\bar{a} \times \bar{b}| = ab \sin\theta$$Therefore, the product of the two cross product magnitudes is:$$|\bar{a} \times \bar{b}| \cdot |\bar{a} \times \bar{b}| = (ab \sin\theta)(ab \sin\theta) = a^{2}b^{2}\sin^{2}\theta$$Adding both results:$$|\bar{a} \cdot \bar{b}|^{2} + |\bar{a} \times \bar{b}| \cdot |\bar{a} \times \bar{b}| = a^{2}b^{2}\cos^{2}\theta + a^{2}b^{2}\sin^{2}\theta$$Factor out $a^{2}b^{2}$:$$= a^{2}b^{2}(\cos^{2}\theta + \sin^{2}\theta)$$Using the identity $\cos^{2}\theta + \sin^{2}\theta = 1$, we get:$$= a^{2}b^{2}$$Answer:$$a^{2}b^{2}$$Hence, the correct option is B.