1
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, $|\vec{a}| = |\vec{b}| = |\vec{c}| = 3$ and $\theta$ is the angle between $\vec{b}$ and $\vec{c}$ then $\tan^2\theta + \cot^2\theta =$
A
$\dfrac{2}{3}$
B
$\dfrac{5}{3}$
C
$\dfrac{8}{3}$
D
$\dfrac{10}{3}$
2
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
The altitude of the parallelopiped, whose coterminous edges are the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} - \hat{k}$, $\vec{c} = \hat{i} + \hat{j} + 3\hat{k}$, where $\vec{a}$, $\vec{b}$ are the sides of the base of parallelopiped, is
A
$\dfrac{2\sqrt{2}}{\sqrt{19}}$ units
B
$\dfrac{\sqrt{2}}{\sqrt{19}}$ units
C
$\dfrac{\sqrt{19}}{\sqrt{2}}$ units
D
$\dfrac{\sqrt{19}}{2\sqrt{2}}$ units
3
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
Let ABCD be a quadrilateral with $\overline{AB} = \vec{a}$, $\overline{AD} = \vec{b}$ and $\overline{AC} = 3\vec{a} + 2\vec{b}$. If its area is $\alpha$ times the area of the parallelogram with AB, AD as adjacent sides, then the value of $\alpha$ is equal to
A
$\dfrac{3}{2}$
B
$\dfrac{5}{2}$
C
$\dfrac{1}{2}$
D
$1$
4
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
Let $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = |\vec{c}|$, $|\vec{c} - \vec{a}| = 2\sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to
A
$\dfrac{3\sqrt{3}}{2}$
B
$\sqrt{\dfrac{3}{2}}$
C
$3\sqrt{\dfrac{3}{2}}$
D
$\dfrac{9\sqrt{3}}{2}$

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