Q72 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors of equal magnitude such that the angle between $\bar{a}$ and $\bar{b}$ is $\alpha$, $\bar{b}$ and $\bar{c}$ is $\beta$, $\bar{c}$ and $\bar{a}$ is $\gamma$.
Then the minimum value of $\cos\alpha + \cos\beta + \cos\gamma$ is ...

  • A
    $\dfrac{1}{2}$
  • B
    $-\dfrac{1}{2}$
  • C
    $\dfrac{3}{2}$
  • Correct answer
    D
    $-\dfrac{3}{2}$

Solution

Let the magnitude of the three vectors be $|\vec{a}| = |\vec{b}| = |\vec{c}| = x$.

The square of the magnitude of the sum of the three vectors is given by:

$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$

Since the square of the magnitude of any vector is always non-negative:

$|\vec{a} + \vec{b} + \vec{c}|^2 \ge 0$

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(|\vec{a}||\vec{b}|\cos\alpha + |\vec{b}||\vec{c}|\cos\beta + |\vec{c}||\vec{a}|\cos\gamma) \ge 0$

Substituting the magnitudes:

$x^2 + x^2 + x^2 + 2(x^2\cos\alpha + x^2\cos\beta + x^2\cos\gamma) \ge 0$

$3x^2 + 2x^2(\cos\alpha + \cos\beta + \cos\gamma) \ge 0$

Since $x^2 > 0$ (assuming non-zero vectors), dividing the entire inequality by $x^2$ gives:

$3 + 2(\cos\alpha + \cos\beta + \cos\gamma) \ge 0$

$2(\cos\alpha + \cos\beta + \cos\gamma) \ge -3$

$\cos\alpha + \cos\beta + \cos\gamma \ge -\dfrac{3}{2}$

Therefore, the minimum value of $\cos\alpha + \cos\beta + \cos\gamma$ is $-\dfrac{3}{2}$.

Hence, the correct answer is option: 'D'.

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