If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

Let $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\vec{b}=2 \hat{i}-\hat{j}-2 \hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and the orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$, then find $|\vec{x}-\vec{y}|$.

$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$, then $\sin (\alpha + \beta) =$

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then:

If $a = i - j$ and $b = i + k$,then a unit vector coplanar with $a$ and $b$ and perpendicular to $a$ is

Given three vectors $a=2 \hat{i}-\hat{j}+\hat{k}$,$b=\hat{i}+2 \hat{j}-\hat{k}$,and $c=\hat{i}+\hat{j}-2 \hat{k}$,a vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{\frac{2}{3}}$ is