MHTCET 18 Apr 2026 Shift 1 Paper

Concept:The square of the magnitude of $\vec{a} + \vec{b} + \vec{c}$ is always non-negative, which gives a lower bound for the sum of cosines.Explanation:Let the common magnitude be $r$.So, $|\vec{a}| = |\vec{b}| = |\vec{c}| = r$.Now, $|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0$.Expanding using dot products:$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{a}\cdot\vec{b} + 2\vec{b}\cdot\vec{c} + 2\vec{c}\cdot\vec{a}$$Since each magnitude is $r$, the sum of squares is $3r^2$.Also, $\vec{a}\cdot\vec{b} = r^2\cos\alpha$, $\vec{b}\cdot\vec{c} = r^2\cos\beta$, and $\vec{c}\cdot\vec{a} = r^2\cos\gamma$.Thus,$$|\vec{a} + \vec{b} + \vec{c}|^2 = 3r^2 + 2r^2(\cos\alpha + \cos\beta + \cos\gamma) \geq 0$$Dividing by $r^2 > 0$:$$3 + 2(\cos\alpha + \cos\beta + \cos\gamma) \geq 0$$Therefore,$$\cos\alpha + \cos\beta + \cos\gamma \geq -\frac{3}{2}$$This minimum is achieved when $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, which occurs when the three equal vectors are mutually inclined at $120^\circ$.Then each cosine equals $\cos 120^\circ = -\frac{1}{2}$.Hence, the minimum sum is:$$-\frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{3}{2}$$So, the correct option is D.Answer:$-\frac{3}{2}$ (Option D)