Q73 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
Let $\bar{a}, \bar{b}, \bar{c}$ be the unit vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and the angle between $\bar{b}$ and $\bar{c}$ is $120^\circ$. If $\bar{a} + \bar{c}$ is perpendicular to $\bar{b} + \bar{c}$ then
- A$(\bar{a} + \bar{c}) \cdot (\bar{b} - \bar{c}) = 1$
- B$(\bar{a} - \bar{c}) \cdot (\bar{b} - \bar{c}) = -2$
- C$(\bar{a} - \bar{c}) \cdot (\bar{b} + \bar{c}) = 1$
- Correct answerD$(\bar{a} - \bar{c}) \cdot (\bar{b} - \bar{c}) = 2$
Solution
Given $|\bar{a}| = |\bar{b}| = |\bar{c}| = 1$
$\bar{a} \cdot \bar{b} = 0$
$\bar{b} \cdot \bar{c} = |\bar{b}||\bar{c}|\cos 120^\circ = -\dfrac{1}{2}$
Since $\bar{a} + \bar{c}$ is perpendicular to $\bar{b} + \bar{c}$, we have:
$(\bar{a} + \bar{c}) \cdot (\bar{b} + \bar{c}) = 0$
$\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} + \bar{c} \cdot \bar{b} + |\bar{c}|^2 = 0$
$0 + \bar{a} \cdot \bar{c} - \dfrac{1}{2} + 1 = 0$
$\bar{a} \cdot \bar{c} = -\dfrac{1}{2}$
Now, evaluating the expression in the options:
$(\bar{a} - \bar{c}) \cdot (\bar{b} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{a} \cdot \bar{c} - \bar{c} \cdot \bar{b} + |\bar{c}|^2$
$= 0 - \left(-\dfrac{1}{2}\right) - \left(-\dfrac{1}{2}\right) + 1$
$= \dfrac{1}{2} + \dfrac{1}{2} + 1 = 2$
$\bar{a} \cdot \bar{b} = 0$
$\bar{b} \cdot \bar{c} = |\bar{b}||\bar{c}|\cos 120^\circ = -\dfrac{1}{2}$
Since $\bar{a} + \bar{c}$ is perpendicular to $\bar{b} + \bar{c}$, we have:
$(\bar{a} + \bar{c}) \cdot (\bar{b} + \bar{c}) = 0$
$\bar{a} \cdot \bar{b} + \bar{a} \cdot \bar{c} + \bar{c} \cdot \bar{b} + |\bar{c}|^2 = 0$
$0 + \bar{a} \cdot \bar{c} - \dfrac{1}{2} + 1 = 0$
$\bar{a} \cdot \bar{c} = -\dfrac{1}{2}$
Now, evaluating the expression in the options:
$(\bar{a} - \bar{c}) \cdot (\bar{b} - \bar{c}) = \bar{a} \cdot \bar{b} - \bar{a} \cdot \bar{c} - \bar{c} \cdot \bar{b} + |\bar{c}|^2$
$= 0 - \left(-\dfrac{1}{2}\right) - \left(-\dfrac{1}{2}\right) + 1$
$= \dfrac{1}{2} + \dfrac{1}{2} + 1 = 2$
Hence, the correct answer is option: 'D'.