Q73 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If a unit vector makes angles $\dfrac{\pi}{4}$ with $\hat{i}$, $\dfrac{\pi}{3}$ with $\hat{j}$ and $\theta \in (0, \pi)$ with $\hat{k}$, then a value of $\theta$ is equal to...
- Correct answerA$\dfrac{\pi}{3}, \dfrac{2\pi}{3}$
- B$\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
- C$\dfrac{\pi}{4}, \dfrac{5\pi}{4}$
- D$\dfrac{5\pi}{12}, \dfrac{7\pi}{12}$
Solution
Let the unit vector be $\vec{a} = \cos\alpha \hat{i} + \cos\beta \hat{j} + \cos\gamma \hat{k}$.
Given $\alpha = \dfrac{\pi}{4}$, $\beta = \dfrac{\pi}{3}$, and $\gamma = \theta$.
Using the property of direction cosines, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
Substituting the given values:
$\cos^2\left(\dfrac{\pi}{4}\right) + \cos^2\left(\dfrac{\pi}{3}\right) + \cos^2\theta = 1$
$\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2 + \cos^2\theta = 1$
$\dfrac{1}{2} + \dfrac{1}{4} + \cos^2\theta = 1$
$\dfrac{3}{4} + \cos^2\theta = 1$
$\cos^2\theta = \dfrac{1}{4}$
$\cos\theta = \pm \dfrac{1}{2}$
Since $\theta \in (0, \pi)$, we get $\theta = \dfrac{\pi}{3}$ or $\theta = \dfrac{2\pi}{3}$.
Answer: $\dfrac{\pi}{3}, \dfrac{2\pi}{3}$
Given $\alpha = \dfrac{\pi}{4}$, $\beta = \dfrac{\pi}{3}$, and $\gamma = \theta$.
Using the property of direction cosines, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
Substituting the given values:
$\cos^2\left(\dfrac{\pi}{4}\right) + \cos^2\left(\dfrac{\pi}{3}\right) + \cos^2\theta = 1$
$\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2 + \cos^2\theta = 1$
$\dfrac{1}{2} + \dfrac{1}{4} + \cos^2\theta = 1$
$\dfrac{3}{4} + \cos^2\theta = 1$
$\cos^2\theta = \dfrac{1}{4}$
$\cos\theta = \pm \dfrac{1}{2}$
Since $\theta \in (0, \pi)$, we get $\theta = \dfrac{\pi}{3}$ or $\theta = \dfrac{2\pi}{3}$.
Answer: $\dfrac{\pi}{3}, \dfrac{2\pi}{3}$
Hence, the correct answer is option: 'A'.