1
COMEDK 2025 Evening Shift
MCQ (Single Correct Answer)
+1
-0
For any vector $\vec{p}$, the value of $\left[2\left\{|\vec{p} \times \hat{\imath}|^2+|\vec{p} \times \hat{\jmath}|^2+|\vec{p} \times \hat{k}|^2\right\}\right]$ is
A
$4|\vec{p}|^2$
B
$2|\vec{p}|^2$
C
$4|\vec{p}|$
D
$2|\vec{p}|$
2
COMEDK 2025 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0
If $|\vec{a}|=2 \sqrt{2}$ and $|\vec{b}|=3$ and angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. If a parallelogram is constructed with adjacent sides $\vec{p}=2 \vec{a}-3 \vec{b}$ and $\vec{q}=\vec{a}+\vec{b}$ then the product of length of both the diagonals is :
A
$12 \sqrt{26}$
B
$6$
C
$60 \sqrt{2}$
D
$18 \sqrt{260}$
3
COMEDK 2025 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0
Position vector of P and Q are $\hat{\imath}+3 \hat{\jmath}-7 \hat{k}$ and $5 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}$ respectively. Then the cosine of the angle between $\overrightarrow{P Q}$ and y -axis is
A
$\frac{4}{\sqrt{162}}$
B
$\frac{5}{\sqrt{162}}$
C
$-\frac{5}{\sqrt{162}}$
D
$-\frac{4}{\sqrt{162}}$
4
COMEDK 2025 Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $a \neq 0$ and $\vec{a} \times \vec{b}=2(\vec{a} \times \vec{c}),|\vec{a}|=|\vec{c}|=1,|\vec{b}|=4$ and $|\vec{b} \times \vec{c}|=\sqrt{15}$ if $\vec{b}-2 \vec{c}=\lambda \vec{a}$ then $\lambda^2$ equals :
A
$-$4
B
16
C
1
D
4
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