JEE Main 2024 (Online) 5th April Morning Shift

MCQ (Single Correct Answer)

-1

The angle between vector $$\vec{Q}$$ and the resultant of $$(2 \vec{Q}+2 \vec{P})$$ and $$(2 \vec{Q}-2 \vec{P})$$ is :

$$ \tan ^{-1}(\mathrm{P} / \mathrm{Q}) $$

0$$^\circ$$

$$ \tan ^{-1} \frac{(2 \vec{Q}-2 \vec{P})}{2 \vec{Q}+2 \vec{P}} $$

$$ \tan ^{-1}(2 Q / \mathrm{P}) $$

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

If two vectors $$\vec{A}$$ and $$\vec{B}$$ having equal magnitude $$R$$ are inclined at angle $$\theta$$, then

$$|\vec{A}+\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)$$

$$|\vec{A}-\vec{B}|=2 R \cos \left(\frac{\theta}{2}\right)$$

$$|\vec{A}-\vec{B}|=\sqrt{2} R \sin \left(\frac{\theta}{2}\right)$$

$$|\vec{A}+\vec{B}|=2 R \sin \left(\frac{\theta}{2}\right)$$

JEE Main 2023 (Online) 15th April Morning Shift

MCQ (Single Correct Answer)

-1

A vector in $x-y$ plane makes an angle of $30^{\circ}$ with $y$-axis. The magnitude of $\mathrm{y}$-component of vector is $2 \sqrt{3}$. The magnitude of $x$-component of the vector will be :

$\sqrt{3}$

$\frac{1}{\sqrt{3}}$

JEE Main 2023 (Online) 11th April Evening Shift

MCQ (Single Correct Answer)

-1

When vector $$\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$$ is subtracted from vector $$\overrightarrow{\mathrm{B}}$$, it gives a vector equal to $$2 \hat{j}$$. Then the magnitude of vector $$\overrightarrow{\mathrm{B}}$$ will be :

$$\sqrt{33}$$

$$\sqrt6$$

$$\sqrt5$$

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