Vector Algebra - Mechanics - Physics Previous Year Questions

$\mathbf{P}$ and $\mathbf{Q}$ are two non-zero vectors inclined to each other at an angle ' $\theta$ '. ' $p$ ' and ' $q$ ' are unit vectors along $\mathbf{P}$ and $\mathbf{Q}$ respectively. The component of $\mathbf{Q}$ in the direction of $\mathbf{Q}$ will be

If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$, then the angle between two vectors is

The resultant $\mathbf{R}$ of $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$. Also $|\mathbf{P}|=|\mathbf{R}|$. The angle between $\mathbf{P}$ and $\mathbf{Q}$ is $\left[\tan 45^{\circ}=1\right]$

A vector $P$ has $X$ and $Y$ components of magnitude 2 units and 4 units respectively. A vector $Q$ along negative $X$-axis has magnitude 6 units. The vector $(\mathbf{Q}-\mathbf{P})$ will be

The vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$ are at right. angles to each other. This is possible under the condition

The \(x, y\) components of vector \(\mathbf{P}\) have magnitudes 1 and 3 and \(x, y\) components of resultant of \(\mathbf{P}\) and \(\mathbf{Q}\) have magnitudes 5 and 6, respectively. What is the magnitude of \(\mathbf{Q}\) ?

The resultant of two vector \(\mathbf{A}\) and \(\mathbf{B}\) is \(\mathbf{C}\). If the magnitude of \(\mathbf{B}\) is doubled, the new resultant vector becomes perpendicular to A. Then, the magnitude of \(\mathbf{C}\) is

Two vectors of same magnitude have a resultant equal to either of the two vectors. The angle between two vectors is

The angle subtended by the vector $A=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}$ with the $X$-axis is

What is the angle between resultant of $A+B$ and $\mathbf{A} \times \mathbf{B}$.

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$. If the magnitude of $\vec{B}$ is doubled, the new resultant vector becomes perpendicular to $\vec{A}$, then the magnitude of $\overrightarrow{\mathrm{C}}$ is

Three vectors are expressed as $\vec{a}=4 \hat{i}-\hat{j}, \vec{b}=-3 \hat{i}+2 \hat{j}$ and $\vec{c}=-\hat{k}$. The unit vector along the direction of sum of these vectors is

If $\vec{A}=\hat{i}+\hat{j}+3 \hat{k}, \vec{B}=-\hat{i}+\hat{j}+4 \hat{k}$ and $\vec{C}=2 \hat{i}-2 \hat{j}-8 \hat{k}$, then the angle between the vectors $\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{Q}}=(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})$ is (in degree)

A unit vector in the direction of resultant vector of $\vec{A}=-2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{B}=\hat{i}+2 \hat{j}-4 \hat{k}$ is

The three vector $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 k$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

Given $\quad \vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

Given $\quad \vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

If $\vec{P}=b \hat{i}+6 \hat{j}+\hat{k} \quad$ and $\quad \vec{Q}=\hat{i}-a \hat{j}+4 \hat{k} \quad$ are perpendicular to each other, also $3 \mathrm{~b}-\mathrm{a}=5$. The value of $a$ and $b$ is

Vector $\vec{A}$ of magnitude $5 \sqrt{3}$ units, another vector $\vec{B}$ of magnitude of 10 units are inclined to each other at an angle of $30^{\circ}$. The magnitude of vector product of the two vectors is $\left[\sin 30^{\circ}=\frac{1}{2}\right]$

The vector sum of two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence forces $\vec{A}$ and $\vec{B}$ are