MHT CET 2026 18th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 18th April Evening Shift

MCQ (Single Correct Answer)

-0

The value of $\theta \in \left(0, \dfrac{\pi}{2}\right)$ for which vectors $\bar{a} = (\sin\theta)\hat{i} + (\cos\theta)\hat{j}$ and $\bar{b} = \hat{i} - \sqrt{3}\hat{j} + 2\hat{k}$ are perpendicular is

$\theta = \dfrac{\pi}{3}$

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

$\theta = \dfrac{\pi}{4}$

$\theta = \dfrac{\pi}{2}$

MHT CET 2026 18th April Evening Shift

MCQ (Single Correct Answer)

-0

If $|\bar{a}| = |\bar{b}| = 1, |\bar{c}| = 2$ and $\bar{a} \times (\bar{a} \times \bar{c}) + \bar{b} = \bar{0}$, then the acute angle between $\bar{a}$ and $\bar{c}$ is $\ldots$

$\dfrac{\pi}{2}$

$\dfrac{\pi}{3}$

$\dfrac{\pi}{4}$

$\dfrac{\pi}{6}$

MHT CET 2026 18th April Evening Shift

MCQ (Single Correct Answer)

-0

The value of $|\bar{a} \cdot \bar{b}|^2 + |\bar{a} \times \bar{b}| \cdot |\bar{a} \times \bar{b}|$ is $\ldots$

$-a^2 b^2$

$a^2 b^2$

$a^2 b^2 \cos\theta$

$a^2 b^2 \sin\theta$

MHT CET 2026 18th April Evening Shift

MCQ (Single Correct Answer)

-0

If $\bar{a} \cdot \bar{b} = \beta$ and $\bar{a} \times \bar{b} = \bar{c}$ then $\bar{a} = $

$\dfrac{\bar{b} \times \bar{c} - \beta\bar{b}}{|\bar{b}|^2}$

$\dfrac{\bar{b} \times \bar{c} - \beta\bar{c}}{|\bar{b}|^2}$

$\dfrac{\bar{b} \times \bar{c} + \beta\bar{b}}{|\bar{b}|^2}$

$\dfrac{\bar{b} \times \bar{c} + \beta\bar{c}}{|\bar{b}|^2}$

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