MHTCET 20 Apr 2026 Shift 2 Paper

Concept:At the point of intersection, the position vectors satisfy both line equations, allowing us to relate $\vec{a}$ and $\vec{b}$ to the given point.Explanation:Let the intersection point be $\vec{p} = 2\hat{i} + 4\hat{j} - 4\hat{k}$.Since $\vec{p}$ lies on both lines,$\vec{p} = \vec{a} + \lambda \vec{b}$ and $\vec{p} = \vec{b} + \mu \vec{a}$.Equating these gives$\vec{a} + \lambda \vec{b} = \vec{b} + \mu \vec{a}$.Rearranging,$(1 - \mu)\vec{a} = (1 - \lambda)\vec{b}$.The lines intersect at a single point, so $\vec{a}$ and $\vec{b}$ are not parallel.Thus, both coefficients must be zero:$1 - \mu = 0$ and $1 - \lambda = 0$.So $\lambda = 1$ and $\mu = 1$.Therefore, $\vec{p} = \vec{a} + \vec{b}$.Now,$|\vec{a} + \vec{b}|^2 = |\vec{p}|^2 = 2^2 + 4^2 + (-4)^2 = 36$.Given $|\vec{a} - \vec{b}| = 4$, so $|\vec{a} - \vec{b}|^2 = 16$.Using the identities:$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$,$|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}$.Subtracting them:$36 - 16 = 4\vec{a}\cdot\vec{b}$.So $20 = 4\vec{a}\cdot\vec{b}$, giving $\vec{a}\cdot\vec{b} = 5$.Answer:$\vec{a} \cdot \vec{b} = 5$, which is option A.