vectors Questions

A force $\vec{F}=(5\hat{i}+3\hat{j}+2\hat{k})\text{N}$ is applied over a particle which displaces it from its origin to the point $\vec{r}=(2\hat{i}-\hat{j})\text{m}$. The work done on the particle in joules is:

If $\vec{A}\times \vec{B}=\vec{B}\times \vec{A}$, then the angle between $\vec{A}$ and $\vec{B}$ is:

A particle has two velocities of equal magnitude $u$ inclined to each other at an angle $\theta$. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $\theta$ is:

The values of $a$ for which the points $A,B,C$ with position vectors $2\hat{i}-\hat{j}+\hat{k}$, $\hat{i}-3\hat{j}-5\hat{k}$ and $a\hat{i}-3\hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $\angle C=\pi /2$ are:

Let $\vec{a},\vec{b}$ and $\vec{c}$ be non-zero vectors such that $(\vec{a}\times \vec{b})\times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}|\vec{a}.$ If $\theta$ is the acute angle between the vectors $\vec{b}$ and $\vec{c},$ then $\sin \theta$ equals

Let $\vec{u},\vec{v},\vec{w}$ be such that $|\vec{u}|=1,|\vec{v}|=2,|\vec{w}|=3.$ If the projection $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v},\vec{w}$ are perpendicular to each other then $|\vec{u}-\vec{v}+\vec{w}|$ equals

If $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\vec{a}+2\vec{b}+3\vec{c},\lambda \vec{b}+4\vec{c}$ and $(2\lambda -1)\vec{c}$ are non-coplanar for

A particle is acted upon by constant forces $4\hat{i}+\hat{j}-3\hat{k}$ and $3\hat{i}+\hat{j}-\hat{k}$ which displace it from a point $\hat{i}+2\hat{j}+3\hat{k}$ to the point $5\hat{i}+4\hat{j}+\hat{k}.$ The work done in standard units by the forces is given by

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$ then $\vec{a}+2\vec{b}+6\vec{c}$ equals

If $(\vec{a}\times \vec{b})\times \vec{c}=\vec{a}\times (\vec{b}\times \vec{c})$, where $\vec{a},\vec{b}$ and $\vec{c}$ are any three vectors such that $\vec{a}\cdot \vec{b}\ne 0$ and $\vec{b}\cdot \vec{c}\ne 0$, then $\vec{a}$ and $\vec{c}$ are:

$ABC$ is a triangle, right-angled at $A$. The resultant of the forces acting along $AB$ and $AC$ with magnitudes $\frac{1}{AB}$ and $\frac{1}{AC}$ respectively is a force along $AD$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. The magnitude of the resultant is:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x$ equals:

If a line makes an angle of $\pi /4$ with the positive directions of each of $x$ -axis and $y$ -axis, then the angle that the line makes with the positive direction of the $z$ -axis is:

If $\hat{u}$ and $\hat{v}$ are unit vectors and $\theta$ is the acute angle between them, then $2\hat{u}\times 3\hat{v}$ is a unit vector for:

The resultant of two forces $P\text{N}$ and $3\text{N}$ is a force of $7\text{N}$. If the direction of $3\text{N}$ force were reversed, the resultant would be $\sqrt{19}\text{N}$. The value of $P$ is:

Let $\vec{u},\vec{v},\vec{w}$ be non-coplanar vectors and $p,q$ are real numbers, then the equality $[3\vec{u}p\vec{v}p\vec{w}]-[p\vec{v}\vec{w}q\vec{u}]-[2\vec{w}q\vec{v}q\vec{u}]=0$ holds for:

Let $ABCD$ be a parallelogram such that $\overrightarrow{AB}=\vec{q}$, $\overrightarrow{AD}=\vec{p}$ and $\angle BAD$ be an acute angle. If $\vec{r}$ is the vector that coincides with the altitude directed from the vertex $B$ to the side $AD$, then $\vec{r}$ is given by:

A particle has an initial velocity $3\hat{i}+4\hat{j}$ and an acceleration of $0.4\hat{i}+0.3\hat{j}$. Its speed after $10\text{s}$ is:

The projections of a vector on the three coordinate axis are $6,-3,2$ respectively. The direction cosines of the vector are:

Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. Then vector $\vec{b}$ satisfying $\vec{a}\times \vec{b}+\vec{c}=\vec{0}$ and $\vec{a}\cdot \vec{b}=3$ is:

If the vectors $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$, $\vec{b}=2\hat{i}+4\hat{j}+\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$ are mutually orthogonal, then $(\lambda ,\mu )=$

The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying: $\vec{b}\times \vec{c}=\vec{b}\times \vec{d}$ and $\vec{a}\cdot \vec{d}=0$. Then the vector $\vec{d}$ is equal to:

If $\vec{a}=\frac{1}{\sqrt{10}}(3\hat{i}+\hat{k})$ and $\vec{b}=\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})$, then the value of $(\overrightarrow{2a}-\vec{b})\cdot [(\vec{a}\times \vec{b})\times (\vec{a}+\overrightarrow{2b})]$ is:

Let $\hat{a}$ and $\hat{b}$ be two unit vectors. If the vectors $\vec{c}=\hat{a}+2\hat{b}$ and $\vec{d}=5\hat{a}-4\hat{b}$ are perpendicular to each other, then the angle between $\hat{a}$ and $\hat{b}$ is:

A particle is moving eastwards with a velocity of $5\text{ m/s}$. In $10\text{ seconds}$ the velocity changes to $5\text{ m/s}$ northwards. The average acceleration in this time is:

The vector $\vec{a}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}=\hat{j}+\hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

The non-zero vectors $\vec{a},\vec{b}$ and $\vec{c}$ are related by $\vec{a}=8\vec{b}$ and $\vec{c}=-7\vec{b}$. Then the angle between $\vec{a}$ and $\vec{c}$ is:

Three charges $-q_1$, $+q_2$ and $-q_3$ are placed as shown in the figure. ($-q_1$ at origin, $+q_2$ at $(b,0)$, $-q_3$ at $(a\cos \theta ,a\sin \theta )$). The $x$ -component of the force on $-q_1$ is proportional to:

$\vec{a},\vec{b},\vec{c}$ are 3 vectors, such that $\vec{a}+\vec{b}+\vec{c}=0$, $|\vec{a}|=1,|\vec{b}|=2,|\vec{c}|=3$, then $\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}$ is equal to:

If $\vec{u},\vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u}+\vec{v}-\vec{w})\cdot [(\vec{u}-\vec{v})\times (\vec{v}-\vec{w})]$ equals:

A tetrahedron has vertices at $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$ and $C(-1,1,2)$. Then the angle between the faces $OAB$ and $ABC$ will be:

If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and vectors $(1,a,a^2)$, $(1,b,b^2)$ and $(1,c,c^2)$ are non-coplanar, then the product $abc$ equals:

Consider points $A,B,C$ and $D$ with position vectors $7\hat{i}-4\hat{j}+7\hat{k}$, $\hat{i}-6\hat{j}+10\hat{k}$, $-\hat{i}-3\hat{j}+4\hat{k}$ and $5\hat{i}-\hat{j}+5\hat{k}$ respectively. Then $ABCD$ is a:

A particle acted on by constant forces $4\hat{i}+\hat{j}-3\hat{k}$ and $3\hat{i}+\hat{j}-\hat{k}$ is displaced from the point $\hat{i}+2\hat{j}+3\hat{k}$ to the point $5\hat{i}+4\hat{j}+\hat{k}$. The total work done by the forces is:

Let $\vec{u}=\hat{i}+\hat{j}$, $\vec{v}=\hat{i}-\hat{j}$ and $\vec{w}=2\hat{i}+3\hat{j}+\hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{u}\cdot \hat{n}=0$ and $\vec{v}\cdot \hat{n}=0$, then $|\vec{w}\cdot \hat{n}|$ is equal to: