Q100 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
The vector $\bar{a} + 3\bar{b}$ is perpendicular to $7\bar{a} - 5\bar{b}$ and the vector $\bar{a} - 4\bar{b}$ is perpendicular to $7\bar{a} - 2\bar{b}$. Then the angle between $\bar{a}$ and $\bar{b}$ is
- A$\dfrac{\pi}{2}$
- B$\dfrac{\pi}{4}$
- C$\dfrac{\pi}{6}$
- Correct answerD$\dfrac{\pi}{3}$
Solution
Given that the vectors are perpendicular, their dot products are zero:
$(\bar{a} + 3\bar{b}) \cdot (7\bar{a} - 5\bar{b}) = 0$
$7|\bar{a}|^2 + 16(\bar{a} \cdot \bar{b}) - 15|\bar{b}|^2 = 0$
Also,
$(\bar{a} - 4\bar{b}) \cdot (7\bar{a} - 2\bar{b}) = 0$
$7|\bar{a}|^2 - 30(\bar{a} \cdot \bar{b}) + 8|\bar{b}|^2 = 0$
Subtracting the second equation from the first, we get:
$46(\bar{a} \cdot \bar{b}) - 23|\bar{b}|^2 = 0$
$\bar{a} \cdot \bar{b} = \dfrac{1}{2}|\bar{b}|^2$
Substituting this into the second equation:
$7|\bar{a}|^2 - 30\left(\dfrac{1}{2}|\bar{b}|^2\right) + 8|\bar{b}|^2 = 0$
$7|\bar{a}|^2 - 15|\bar{b}|^2 + 8|\bar{b}|^2 = 0$
$7|\bar{a}|^2 - 7|\bar{b}|^2 = 0 \Rightarrow |\bar{a}| = |\bar{b}|$
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. Then,
$\cos \theta = \dfrac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \dfrac{\dfrac{1}{2}|\bar{b}|^2}{|\bar{b}|^2} = \dfrac{1}{2}$
$\theta = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
$(\bar{a} + 3\bar{b}) \cdot (7\bar{a} - 5\bar{b}) = 0$
$7|\bar{a}|^2 + 16(\bar{a} \cdot \bar{b}) - 15|\bar{b}|^2 = 0$
Also,
$(\bar{a} - 4\bar{b}) \cdot (7\bar{a} - 2\bar{b}) = 0$
$7|\bar{a}|^2 - 30(\bar{a} \cdot \bar{b}) + 8|\bar{b}|^2 = 0$
Subtracting the second equation from the first, we get:
$46(\bar{a} \cdot \bar{b}) - 23|\bar{b}|^2 = 0$
$\bar{a} \cdot \bar{b} = \dfrac{1}{2}|\bar{b}|^2$
Substituting this into the second equation:
$7|\bar{a}|^2 - 30\left(\dfrac{1}{2}|\bar{b}|^2\right) + 8|\bar{b}|^2 = 0$
$7|\bar{a}|^2 - 15|\bar{b}|^2 + 8|\bar{b}|^2 = 0$
$7|\bar{a}|^2 - 7|\bar{b}|^2 = 0 \Rightarrow |\bar{a}| = |\bar{b}|$
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. Then,
$\cos \theta = \dfrac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \dfrac{\dfrac{1}{2}|\bar{b}|^2}{|\bar{b}|^2} = \dfrac{1}{2}$
$\theta = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Hence, the correct answer is option: 'D'.