Q97 / 150 · MHT CET 2026 (13 April Shift 1) · Mathematics

Let $\bar{a}, \bar{b}, \bar{c}$ be three vectors having magnitudes 1, 1 and 2 respectively. If $\bar{a} \times (\bar{a} \times \bar{c}) + \bar{b} = \bar{0}$, then the acute angle between $\bar{a}$ and $\bar{c}$ is

  • A
    $\dfrac{\pi}{4}$
  • Correct answer
    B
    $\dfrac{\pi}{6}$
  • C
    $\dfrac{\pi}{3}$
  • D
    $\dfrac{\pi}{8}$

Solution

Given $\bar{a} \times (\bar{a} \times \bar{c}) + \bar{b} = \bar{0}$

Using the vector triple product expansion:

$(\bar{a} \cdot \bar{c})\bar{a} - (\bar{a} \cdot \bar{a})\bar{c} + \bar{b} = \bar{0}$

Since $|\bar{a}| = 1$, $\bar{a} \cdot \bar{a} = 1$.

$(\bar{a} \cdot \bar{c})\bar{a} - \bar{c} + \bar{b} = \bar{0}$

$\bar{b} = \bar{c} - (\bar{a} \cdot \bar{c})\bar{a}$

Squaring both sides:

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

$|\bar{b}|^2 = |\bar{c}|^2 - (\bar{a} \cdot \bar{c})^2$

Substituting the given magnitudes $|\bar{a}| = 1$, $|\bar{b}| = 1$, and $|\bar{c}| = 2$:

$1^2 = 2^2 - (|\bar{a}||\bar{c}|\cos\theta)^2$

$1 = 4 - (1 \times 2 \times \cos\theta)^2$

$1 = 4 - 4\cos^2\theta$

$4\cos^2\theta = 3$

$\cos^2\theta = \dfrac{3}{4}$

$\cos\theta = \pm\dfrac{\sqrt{3}}{2}$

For the acute angle, $\cos\theta = \dfrac{\sqrt{3}}{2}$

$\theta = \dfrac{\pi}{6}$

Answer: $\dfrac{\pi}{6}$

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

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